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# Moles of Sulfur monochloride ## sulfur monochloride: convert moles to volume and weight ### Volume of 1 mole of Sulfur monochloride centimeter³ 79.9 milliliter 79.9 foot³ 0 oil barrel 0 Imperial gallon 0.02 US cup 0.34 inch³ 4.88 US fluid ounce 2.7 liter 0.08 US gallon 0.02 meter³ 7.99 × 10-5 US pint 0.17 metric cup 0.32 US quart 0.08 metric tablespoon 5.33 US tablespoon 5.4 metric teaspoon 15.98 US teaspoon 16.21 ### Weight of 1 mole of Sulfur monochloride carat 675.18 ounce 4.76 gram 135.04 pound 0.3 kilogram 0.14 tonne 0 milligram 135 036 ### The entered amount of Sulfur monochloride in various units of amount of substance centimole 100 micromole 1 000 000 decimole 10 millimole 1 000 gigamole 1 × 10-9 mole 1 kilogram-mole 0 nanomole 1 000 000 000 kilomole 0 picomole 1 000 000 000 000 megamole 1 × 10-6 pound-mole 0 #### Foods, Nutrients and Calories ROTISSERIE SEASONED CHICKEN BREAST STRIPS, SMOKE, UPC: 034695122617 contain(s) 129 calories per 100 grams (≈3.53 ounces)  [ price ] 13659 foods that contain Folate, food.  List of these foods starting with the highest contents of Folate, food and the lowest contents of Folate, food #### Gravels, Substances and Oils CaribSea, Freshwater, Instant Aquarium, Crystal River weighs 1 521.75 kg/m³ (94.99975 lb/ft³) with specific gravity of 1.52175 relative to pure water.  Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical  or in a rectangular shaped aquarium or pond  [ weight to volume | volume to weight | price ] Potassium peroxymonosulfate [2KHSO5 ⋅ KHSO4 ⋅ K2SO4 ] weighs 2 350 kg/m³ (146.70571 lb/ft³)  [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ] Volume to weightweight to volume and cost conversions for Linseed oil with temperature in the range of 10°C (50°F) to 140°C (284°F) #### Weights and Measurements The ounce (oz) is a non-metric unit of mass The units of data measurement were introduced to manage and operate digital information. oz t/metric c to lb/mm³ conversion table, oz t/metric c to lb/mm³ unit converter or convert between all units of density measurement. #### Calculators Cube calculator. Compute the volume of a cube and its surface area

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### Showing videos from Eddie Woo with a total of 4,051 videos 04:31 Max/Min Question: Cutting a Wire in Two (2 of 2: Finding the minimum) 1y9m ago 10:08 Max/Min Question: Cutting a Wire in Two (1 of 2: Setting up the equations) 1y9m ago 06:01 The Power Curves (2 of 2: Considering higher powers) 1y9m ago 07:20 The Power Curves (1 of 2: Investigating through tables of values) 1y9m ago 09:54 Normal Distribution - HSC Question 1y9m ago 05:47 Maximum/Minimum with Quadratics (2 of 2: Completing the square) 1y9m ago 08:30 Maximum/Minimum with Quadratics (1 of 2: Axis of symmetry) 1y9m ago 10:08 Graphs of Logarithms (3 of 3: Visualising the shape) 1y9m ago 09:16 Graphs of Logarithms (2 of 3: Investigating near x=0) 1y9m ago 07:59 Graphs of Logarithms (1 of 3: Considering integer values) 1y9m ago 08:37 Equations Reducible to Quadratics (Changing Logarithm Bases) 1y9m ago 05:26 Solving Equations with Radical Functions (2 of 2: Final solution) 1y9m ago 10:58 Solving Equations with Radical Functions (1 of 2: Initial steps & discussion) 1y9m ago 10:00 Equation of a Parabola - Vertex Form (Completing the square) 1y9m ago 06:30 Equation of a Translated Circle 1y9m ago 06:10 Y10 Review Questions (Solving quadratic equations) 1y9m ago 11:40 Proof by Contradiction (2 of 2: Infinite primes) 1y9m ago 08:34 Proof by Contradiction (1 of 2: How does it work?) 1y9m ago 06:58 Introduction to Quadratic Theory (2 of 2: Completing the square) 1y9m ago 06:45 Introduction to Quadratic Theory (1 of 2: Why we care about quadratics) 1y9m ago 06:42 The Circle (2 of 2: Constructing the Cartesian equation) 1y9m ago 07:34 The Circle (1 of 2: Starting with a verbal definition) 1y9m ago 08:16 Y10 Review Questions (Simple graphs) 1y9m ago 13:20 The Hyperbola (2 of 2: Understanding the graph & its behaviour) 1y9m ago 10:19 The Hyperbola (1 of 2: Review of earlier graphs) 1y9m ago 12:28 Intercept Properties - Circle Geometry (3 of 3: Tangent & secant) 1y9m ago 05:14 Intercept Properties - Circle Geometry (2 of 3: Secants from a point) 1y9m ago 05:59 Intercept Properties - Circle Geometry (1 of 3: Intersecting chords) 1y9m ago 09:53 Angle in the Alternate Segment - Circle Geometry (3 of 3: Conclusion and overall summary) 1y9m ago 09:39 Angle in the Alternate Segment - Circle Geometry (2 of 3: Proof) 1y9m ago 07:15 Angle in the Alternate Segment - Circle Geometry (1 of 3: Setup) 1y9m ago 09:21 Tangent Properties - Circle Geometry (Tangents from an external point are equal) 1y9m ago 09:21 Tangent Properties - Circle Geometry (Tangent is perpendicular to radius) 1y9m ago 06:53 Enrichment Problem: \$1,000,000 in Specific Denominations (2 of 2: Solution) 1y9m ago 08:19 Enrichment Problem: \$1,000,000 in Specific Denominations (1 of 2: Setup) 1y9m ago 08:40 Enrichment Problem: Arranging Counters on a Plane 1y9m ago 10:51 Cubic Curves 1y9m ago 11:04 Y10 Review Questions (Graphing parabolas) 1y9m ago 10:19 Angle Properties - Circle Geometry (Angles in the same segment) 1y9m ago 07:21 Angle Properties - Circle Geometry (Second proof of angle at the centre/circumference) 1y9m ago 08:37 HSC Geometry Question (2 of 2: Proving the inequality) 1y9m ago 07:50 HSC Geometry Question (1 of 2: Proving & using triangle similarity) 1y9m ago 11:12 Prelim Mathematics Ext 1 Quiz (Inequality regions, solving trigonometric equations) 1y9m ago 07:19 Angle Properties - Circle Geometry (Angle at the centre/circumference) 1y9m ago 09:40 Angle Properties - Circle Geometry (Angle in a semicircle) 1y9m ago 13:21 Chord Properties (Circle Geometry) 1y9m ago 13:20 Components of a Circle 1y9m ago 08:34 Ratios of Intercepts on Transversals (2 of 2: Constructing the proof) 1y9m ago 11:37 Ratios of Intercepts on Transversals (1 of 2: Setting up the problem) 1y9m ago 11:20 Introduction to Parabolas (2 of 2: Characteristics & translations) 1y9m ago 10:34 Introduction to Parabolas (1 of 2: Investigating the shape) 1y9m ago 07:29 Y10 Review Questions (Solving an inequality, rationalising a denominator) 1y9m ago 09:38 Inequality Proofs (Example 5 of 5: Investigating a large product) 1y9m ago 05:25 Inequality Proofs (Example 4 of 5: Arithmetic & Geometric Means) 1y9m ago 06:15 Inequality Proofs (Example 3 of 5: Is 2^300 bigger or 3^200?) 1y9m ago 05:42 Inequality Proofs (Example 2 of 5: Which number is largest?) 1y10m ago 11:07 Inequality Proofs (Example 1 of 5: How many positive integer solutions?) 1y10m ago 09:14 Financial Expectation (2 of 2: When the game isn't free) 1y10m ago 04:44 Financial Expectation (1 of 2: Introduction) 1y10m ago 11:17 Triangle Similarity (quick review) 1y10m ago 12:05 Expected Frequency 1y10m ago 12:53 Quadrilaterals (Relationships, Properties & Sufficiency Conditions) 1y10m ago 12:56 Triangle Congruence (quick review) 1y10m ago 06:40 Probability Trees (2 of 2: How to read them) 1y10m ago 07:36 Probability Trees (1 of 2: How to construct them) 1y10m ago 08:02 Number Bases (5 of 5: Conversion to binary & hexadecimal) 1y10m ago 06:49 Number Bases (4 of 5: Converting between bases) 1y10m ago 07:10 Number Bases (3 of 5: What is the "base" of a number system?) 1y10m ago 08:36 Number Bases (2 of 5: Investigating decimal expansions) 1y10m ago 10:15 Number Bases (1 of 5: Exploring numbers through a spreadsheet) 1y10m ago 04:44 Variation (3 of 3: Inverse example) 1y10m ago 04:53 Variation (2 of 3: Direct example) 1y10m ago

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# Long Division Calculator That Shows and Explains Each Step This calculator will divide one number (dividend) by another number (divisor) using the long division method, and show and explain each step. The calculator will accommodate divisors and dividends containing decimal points and will give the remainder in both the whole number and the decimal format. Plus, unlike other online long division calculators, the calculated results will list the 1-9 multiples of the divisor to help you learn your factors, and will also report the detection of patterns which may indicate the result is a recurring decimal. Note that you are looking to multiply decimals, integers, or whole numbers, check out the Long Multiplication Calculator. ## Long Division Calculator Divide one number by another and see how the result is arrived at using long division. Special Instructions #### Selected Data Record: A Data Record is a set of calculator entries that are stored in your web browser's Local Storage. If a Data Record is currently selected in the "Data" tab, this line will list the name you gave to that data record. If no data record is selected, or you have no entries stored for this calculator, the line will display "None". DataData recordData recordSelected data record: None Divisor:Divisor:Divisor:Divisor: #### Divisor: Enter the divisor in the left-hand field on the line below (number greater than zero). The divisor is the number being divided into another number. If the divisor you enter is greater than the dividend, the result will be a decimal number less than 1. Dividend:Dividend:Dividend:Dividend: #### Dividend: Enter the dividend in the right-hand field on the line below (number greater than zero). The dividend is the number being divided by another number. # Divisor # Dividend Quotient:Division quotient:Long division quotient:Long division quotient: #### Long division quotient: This is the result the long division calculator stopped at. Please double check this result with the result on the next line. Calculated:Calculated quotient:Calculated division quotient:Calculated division quotient: #### Calculated division quotient: This is the result you would likely get if you used a standard calculator for the division. This result is provided as a means to check the long division result for accuracy. Note that if the result contains more than 16 digits the last digit(s) is likely rounded. Whole num:Whole number:Whole number portion of quotient:Whole number portion of quotient: #### Whole number portion of quotient: This is the whole number portion of the quotient. The quotient is the result of dividing one number by another. Remainder:Remainder:Remainder:Remainder: #### Remainder: This is what is left after multiplying the whole number portion of the quotient by the divisor, and then subtracting that result from the dividend. If you would like to save the current entries to the secure online database, tap or click on the Data tab, select "New Data Record", give the data record a name, then tap or click the Save button. To save changes to previously saved entries, simply tap the Save button. Please select and "Clear" any data records you no longer need. ## Learn ### How to do long division. To show you how to do long division I will give the actual calculator results involving four common long division scenarios: • Dividing whole numbers with no remainder. • Dividing whole numbers with whole remainder. • Dividing decimal numbers with no remainder. • Dividing decimal numbers with decimal remainder. To follow the steps I took as I proceeded through each long division example, tap the info (i) icon on each row of the division process. #### Dividing Whole Numbers With No Remainder « Quotient Divisor » 3 1443 « Dividend Multiplesof 3 1 3 2 6 3 9 4 12 5 15 6 18 7 21 8 24 9 27 1 2 3 4 0 4 8 1 3 1 4 4 3 Since 3 is greater than 1 you enter a zero in the 1st answer position and then combine the digit 1 with the digit from the next position in the dividend (4 in column #2) in order to form a number that is greater than or equal to the 3. That new number is 14.Since the greatest multiple of 3 that divides into 14 without going over is 4 (4 x 3 = 12), you enter 4 in the 2nd answer position and 12 in the subtraction row.Since 14 minus 12 leaves a remainder of 2 you enter 2 on the next line. Then move the next dividend digit (4 in column #3) down to that line to form the new subtrahend of 24. - 1 2 Since the greatest multiple of 3 that divides into 24 without going over is 8 (8 x 3 = 24), you enter 8 in the 3rd answer position and 24 in the subtraction row.Since 24 minus 24 leaves a remainder of 0 you enter 0 on the next line. Then move the next dividend digit (3 in column #4) down to that line to form the new subtrahend of 03. 2 4 - 2 4 Since the greatest multiple of 3 that divides into 3 without going over is 1 (1 x 3 = 3), you enter 1 in the 4th answer position and 3 in the subtraction row.Since 3 minus 3 leaves a remainder of 0 you enter 0 on the next line. 0 3 - 3 Since all dividend digits have been moved down, and the remainder is zero, the problem is solved. 0 - Since 3 x 481 does equal 1443, our solution checks out. #### Dividing Whole Numbers With Remainder « Quotient Divisor » 5 123 « Dividend Multiplesof 5 1 5 2 10 3 15 4 20 5 25 6 30 7 35 8 40 9 45 1 2 3 0 2 4 5 1 2 3 Since 5 is greater than 1 you enter a zero in the 1st answer position and then combine the digit 1 with the digit from the next position in the dividend (2 in column #2) in order to form a number that is greater than or equal to the 5. That new number is 12.Since the greatest multiple of 5 that divides into 12 without going over is 2 (2 x 5 = 10), you enter 2 in the 2nd answer position and 10 in the subtraction row.Since 12 minus 10 leaves a remainder of 2 you enter 2 on the next line. Then move the next dividend digit (3 in column #3) down to that line to form the new subtrahend of 23. - 1 0 Since the greatest multiple of 5 that divides into 23 without going over is 4 (4 x 5 = 20), you enter 4 in the 3rd answer position and 20 in the subtraction row.Since 23 minus 20 leaves a remainder of 3 you enter 3 on the next line. 2 3 - 2 0 Since all dividend digits have been moved down and you are left with the number 3, the remainder is 3. 3 Since 5 x 24 + 3 does equal 123, our solution checks out. #### Dividing Decimal Numbers With No Remainder « Quotient Divisor » 5.5 220 « Dividend Multiplesof 55 1 55 2 110 3 165 4 220 5 275 6 330 7 385 8 440 9 495 1 2 3 4 0 0 4 0 The first step in solving the problem is to get rid of the decimal point in the divisor 5.5.To do that, you move the decimal point 1 place to the right in both the divisor 5.5 and the dividend 220.Moving the decimal point 1 place to the right in the divisor changes it to 55.Since the dividend (220) has no decimal places, you simply add a zero for each decimal place you moved in the divisor. This leaves you with a new dividend of 2200.Note that a vertical red line indicates the position of the decimal point within the quotient (if applicable). 5 5 2 2 0 0 Since 55 is greater than 2 you enter a zero in the 1st quotient position and then combine the digit 2 with the digit from the next position in the dividend (2 in column #2) in an attempt to form a number that is greater than or equal to the 55. That new number is 22.Since 55 is greater than 22 you enter a zero in the 2nd quotient position and then combine the digits 22 with the digit from the next position in the dividend (0 in column #3) in an attempt to form a number that is greater than or equal to the 55. That new number is 220.Since the greatest multiple of 55 that divides into 220 without going over is 4 (4 x 55 = 220), you enter 4 in the 3rd quotient position and 220 in the subtraction row.Since 220 minus 220 leaves a remainder of 0 you enter 0 on the next line. Then move the next dividend digit (0 in column #4) down to that line to form the new subtrahend of 00. - 2 2 0 The problem appears to be solved. 0 - Since 5.5 x 40 does equal 220, our solution checks out. #### Dividing Decimal Numbers With Decimal Remainder « Quotient Divisor » 3.33 99.99 « Dividend Multiples of 333 1 333 2 666 3 999 4 1332 5 1665 6 1998 7 2331 8 2664 9 2997 1 2 3 4 5 6 7 8 9 10 11 12 13 0 0 3 0 0 2 7 0 2 7 0 2 7 The first step in solving the problem is to get rid of the decimal point in the divisor 3.33.To do that, you move the decimal point 2 places to the right in both the divisor 3.33 and the dividend 99.99.Moving the decimal point 2 places to the right in the divisor changes it to 333.Since the dividend (99.99) already has 2 decimal places, you simply move the decimal point 2 places to the right. This leaves you with a new dividend of 9999.Next, continue to add zeros to the dividend as needed until you either solve the division, or you reach the desired number of decimal places.Note that a vertical red line indicates the position of the decimal point within the quotient (if applicable). 3 3 3 9 9 9 9 0 0 0 0 0 0 0 0 0 Since 333 is greater than 9 you enter a zero in the 1st quotient position and then combine the digit 9 with the digit from the next position in the dividend (9 in column #2) in an attempt to form a number that is greater than or equal to the 333. That new number is 99.Since 333 is greater than 99 you enter a zero in the 2nd quotient position and then combine the digits 99 with the digit from the next position in the dividend (9 in column #3) in an attempt to form a number that is greater than or equal to the 333. That new number is 999.Since the greatest multiple of 333 that divides into 999 without going over is 3 (3 x 333 = 999), you enter 3 in the 3rd quotient position and 999 in the subtraction row.Since 999 minus 999 leaves a remainder of 0 you enter 0 on the next line. Then move the next dividend digit (9 in column #4) down to that line to form the new subtrahend of 09. - 9 9 9 Since 333 is greater than 9 you enter a zero in the 4th quotient position and then combine the digits 9 with the digit from the next position in the dividend (0 in column #5) in an attempt to form a number that is greater than or equal to the 333. That new number is 90.Since 333 is greater than 90 you enter a zero in the 5th quotient position and then combine the digits 90 with the digit from the next position in the dividend (0 in column #6) in an attempt to form a number that is greater than or equal to the 333. That new number is 900.Since the greatest multiple of 333 that divides into 900 without going over is 2 (2 x 333 = 666), you enter 2 in the 6th quotient position and 666 in the subtraction row.Since 900 minus 666 leaves a remainder of 234 you enter 234 on the next line. Then move the next dividend digit (0 in column #7) down to that line to form the new subtrahend of 2340. 0 9 0 0 - 6 6 6 Since the greatest multiple of 333 that divides into 2340 without going over is 7 (7 x 333 = 2331), you enter 7 in the 7th quotient position and 2331 in the subtraction row.Since 2340 minus 2331 leaves a remainder of 9 you enter 9 on the next line. Then move the next dividend digit (0 in column #8) down to that line to form the new subtrahend of 90. 2 3 4 0 - 2 3 3 1 Since 333 is greater than 90 you enter a zero in the 8th quotient position and then combine the digits 90 with the digit from the next position in the dividend (0 in column #9) in an attempt to form a number that is greater than or equal to the 333. That new number is 900.Since the greatest multiple of 333 that divides into 900 without going over is 2 (2 x 333 = 666), you enter 2 in the 9th quotient position and 666 in the subtraction row.Since 900 minus 666 leaves a remainder of 234 you enter 234 on the next line. Then move the next dividend digit (0 in column #10) down to that line to form the new subtrahend of 2340. 9 0 0 - 6 6 6 Since the greatest multiple of 333 that divides into 2340 without going over is 7 (7 x 333 = 2331), you enter 7 in the 10th quotient position and 2331 in the subtraction row.Since 2340 minus 2331 leaves a remainder of 9 you enter 9 on the next line. Then move the next dividend digit (0 in column #11) down to that line to form the new subtrahend of 90. 2 3 4 0 - 2 3 3 1 Since 333 is greater than 90 you enter a zero in the 11th quotient position and then combine the digits 90 with the digit from the next position in the dividend (0 in column #12) in an attempt to form a number that is greater than or equal to the 333. That new number is 900.Since the greatest multiple of 333 that divides into 900 without going over is 2 (2 x 333 = 666), you enter 2 in the 12th quotient position and 666 in the subtraction row.Since 900 minus 666 leaves a remainder of 234 you enter 234 on the next line. Then move the next dividend digit (0 in column #13) down to that line to form the new subtrahend of 2340. 9 0 0 - 6 6 6 Since the greatest multiple of 333 that divides into 2340 without going over is 7 (7 x 333 = 2331), you enter 7 in the 13th quotient position and 2331 in the subtraction row.Since 2340 minus 2331 leaves a remainder of 9 you enter 9 on the next line. 2 3 4 0 - 2 3 3 1 It appears the calculator ran out of room before it could complete the long division. 9 - Since 3.33 x 30.027027027 does not equal 99.99, either the calculator ran out of room before the long division was completed, the quotient contains a recurring decimal, or there is a rounding issue between the calculated result and the long division result. In this case it appears to be a repeating decimal. I hope I"ve managed to help you to understand how to perform long division. If not, please use the feedback form beneath the calculator to let me know what I have left out. Move the slider to left and right to adjust the calculator width. Note that the Help and Tools panel will be hidden when the calculator is too wide to fit both on the screen. Moving the slider to the left will bring the instructions and tools panel back into view. Also note that some calculators will reformat to accommodate the screen size as you make the calculator wider or narrower. If the calculator is narrow, columns of entry rows will be converted to a vertical entry form, whereas a wider calculator will display columns of entry rows, and the entry fields will be smaller in size ... since they will not need to be "thumb friendly".

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TOPICS # Finite Difference The finite difference is the discrete analog of the derivative. The finite forward difference of a function is defined as (1) and the finite backward difference as (2) The forward finite difference is implemented in the Wolfram Language as DifferenceDelta[f, i]. If the values are tabulated at spacings , then the notation (3) is used. The th forward difference would then be written as , and similarly, the th backward difference as . However, when is viewed as a discretization of the continuous function , then the finite difference is sometimes written (4) (5) where denotes convolution and is the odd impulse pair. The finite difference operator can therefore be written (6) An th power has a constant th finite difference. For example, take and make a difference table, (7) The column is the constant 6. Finite difference formulas can be very useful for extrapolating a finite amount of data in an attempt to find the general term. Specifically, if a function is known at only a few discrete values , 1, 2, ... and it is desired to determine the analytical form of , the following procedure can be used if is assumed to be a polynomial function. Denote the th value in the sequence of interest by . Then define as the forward difference , as the second forward difference , etc., constructing a table as follows (8) (9) (10) (11) Continue computing , , etc., until a 0 value is obtained. Then the polynomial function giving the values is given by (12) (13) When the notation , , etc., is used, this beautiful equation is called Newton's forward difference formula. To see a particular example, consider a sequence with first few values of 1, 19, 143, 607, 1789, 4211, and 8539. The difference table is then given by (14) Reading off the first number in each row gives , , , , . Plugging these in gives the equation (15) (16) which indeed fits the original data exactly. Formulas for the derivatives are given by (17) (18) (19) (20) (21) (22) (23) (24) (25) (26) (27) (Beyer 1987, pp. 449-451; Zwillinger 1995, p. 705). Formulas for integrals of finite differences (28) are given by Beyer (1987, pp. 455-456). Finite differences lead to difference equations, finite analogs of differential equations. In fact, umbral calculus displays many elegant analogs of well-known identities for continuous functions. Common finite difference schemes for partial differential equations include the so-called Crank-Nicolson, Du Fort-Frankel, and Laasonen methods. Backward Difference, Bessel's Finite Difference Formula, Derivative, Difference Equation, Difference Table, Everett's Formula, Finite Element Method, Forward Difference, Gauss's Backward Formula, Gauss's Forward Formula, Interpolation, Jackson's Difference Fan, Newton's Backward Difference Formula, Newton-Cotes Formulas, Newton's Divided Difference Interpolation Formula, Newton's Forward Difference Formula, Quotient-Difference Table, Recurrence Equation, Steffenson's Formula, Stirling's Finite Difference Formula, Umbral Calculus ## Explore with Wolfram|Alpha More things to try: ## References Abramowitz, M. and Stegun, I. A. (Eds.). "Differences." §25.1 in Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover, pp. 877-878, 1972.Beyer, W. H. CRC Standard Mathematical Tables, 28th ed. Boca Raton, FL: CRC Press, pp. 429-515, 1987.Boole, G. and Moulton, J. F. A Treatise on the Calculus of Finite Differences, 2nd rev. ed. New York: Dover, 1960.Conway, J. H. and Guy, R. K. "Newton's Useful Little Formula." In The Book of Numbers. New York: Springer-Verlag, pp. 81-83, 1996.Fornberg, B. "Calculation of Weights in Finite Difference Formulas." SIAM Rev. 40, 685-691, 1998.Iyanaga, S. and Kawada, Y. (Eds.). "Interpolation." Appendix A, Table 21 in Encyclopedic Dictionary of Mathematics. Cambridge, MA: MIT Press, pp. 1482-1483, 1980.Jordan, C. Calculus of Finite Differences, 3rd ed. New York: Chelsea, 1965.Levy, H. and Lessman, F. Finite Difference Equations. New York: Dover, 1992.Milne-Thomson, L. M. The Calculus of Finite Differences. London: Macmillan, 1951.Richardson, C. H. An Introduction to the Calculus of Finite Differences. New York: Van Nostrand, 1954.Spiegel, M. Calculus of Finite Differences and Differential Equations. New York: McGraw-Hill, 1971.Stirling, J. Methodus differentialis, sive tractatus de summation et interpolation serierum infinitarium. London, 1730. English translation by Holliday, J. The Differential Method: A Treatise of the Summation and Interpolation of Infinite Series. 1749.Tweddle, C. James Stirling: A Sketch of His Life and Works Along with his Scientific Correspondence. Oxford, England: Oxford University Press, pp. 30-45, 1922.Weisstein, E. W. "Books about Finite Difference Equations." http://www.ericweisstein.com/encyclopedias/books/FiniteDifferenceEquations.html.Zwillinger, D. (Ed.). "Difference Equations" and "Numerical Differentiation." §3.9 and 8.3.2 in CRC Standard Mathematical Tables and Formulae. Boca Raton, FL: CRC Press, pp. 228-235 and 705-705, 1995. ## Referenced on Wolfram|Alpha Finite Difference ## Cite this as: Weisstein, Eric W. "Finite Difference." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/FiniteDifference.html

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# Thread: What does this line do? 1. ## What does this line do? Hi, just reading through some code and got a little stuck as to the format in which this value is writing to a string... Code: ```char string[128]; int i; for (i=0; i<10;i++){ printf("%08X\n", i); }``` I don't get what the output would be when doing %08x rather than just %x, and I get that by doing %x, you would print the hexidecimal value for at each loop. Thanks! 2. %08x means "fill with zeros to make it 8 digits, in hex". Similarly, you can use %8x, in which case it's filled with spaces instead of zeros, and %-8x would mean "fill with spaces AFTER the number to make it 8 positions". I don't think "%-08x" does anything sensible. (That is, it's probably no different from "%08x" - but it may behave strangely different from bugginess in any given imlementation). -- Mats 3. >I don't get what the output would be Run it and take a look then. Really. It's just a printf statement. 4. Ok... so if i = 2, would output be: 20000000 or would it be 00000002 5. Originally Posted by bobthebullet990 Ok... so if i = 2, would output be: 20000000 or would it be 00000002 What do you think makes most sense (try it and see if you are right. you may want to count a tiny bit higher than 9 to make it show that it's hex!) -- Mats 6. I would love to try it! ...would have saved me having to post, but I'm on a windows machine with no C compiler!!!! 7. Well.. Im guessing it would pad before the digit! as padding after would be stupid! ...So i'm guessing that output would be: 00000000 ... 00000009 0000000A 0000000B .... 8. Originally Posted by bobthebullet990 I would love to try it! ...would have saved me having to post, but I'm on a windows machine with no C compiler!!!! http://www.compilers.net/Dir/Free/Compilers/CCpp.htm

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Home > English > Class 9 > Maths > Chapter > Matrices > A={:[(-5,-3,4),(3,2,-4)]:}andB... # A={:[(-5,-3,4),(3,2,-4)]:}andB={:[(-4,5,-2),(3,1,5)]:}, then find X such that 3A-2B+X=0. Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams. Updated On: 19-10-2020 Apne doubts clear karein ab Whatsapp par bhi. Try it now. Watch 1000+ concepts & tricky questions explained! 9.5 K+ 4.2 K+ Text Solution {:[(7,19,16),(3,4,22)]:}{:[(-7,-19,16),(3,4,22)]:}{:[(7,19,-16),(-3,-4,22)]:}{:[(7,19,16),(-3,-4,22)]:} C Solution : (i) X=2B-3A <br> (ii) 3A-2B+X=0rArrX=2B-3A. <br> (iii) Multiply each element of B with 2 to get 2B. <br> (iv) Multiply each element of A with 3 to get 3B. Image Solution 8484899 4.3 K+ 85.1 K+ 4:08 51234380 39.2 K+ 41.6 K+ 3:21 40251803 2.2 K+ 44.3 K+ 3:49 40251690 700+ 14.8 K+ 1:51 40251754 900+ 18.5 K+ 2:52 92138863 2.3 K+ 18.9 K+ 4:01 8485062 4.5 K+ 90.5 K+ 6:43 40251790 3.7 K+ 9.2 K+ 2:03 51234379 2.1 K+ 41.9 K+ 3:11 40251737 500+ 10.3 K+ 2:07 40251762 2.0 K+ 40.4 K+ 5:54 40251694 12.4 K+ 14.1 K+ 1:55 40251794 1.0 K+ 20.1 K+ 2:19 52004993 6.1 K+ 63.4 K+ 1:45 65361 2.0 K+ 40.3 K+ 7:21

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# Question: Lorkay Seidens Inc just borrowed 25 000 The loan is to Lorkay Seidens Inc. just borrowed \$25,000. The loan is to be repaid in equal installments at the end of each of the next five years, and the interest rate is 10 percent. a. Set up an amortization schedule for the loan. b. How large must each annual payment be if the loan is for \$50,000? Assume that the interest rate remains at 10 percent and that the loan is paid off over five years. c. How large must each payment be if the loan is for \$50,000, the interest rate is 10 percent, and the loan is paid off in equal installments at the end of each of the next 10 years? This loan is for the same amount as the loan in part (b), but the payments are spread out over twice as many periods. Why are these payments not half as large as the payments on the loan in part (b)? View Solution: Sales0 Views286

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London, England interview questions | Glassdoor.co.uk # Interview questions in London, England Bloomberg L.P. Interviews in London www.bloomberg.com /  HQ: New York, NY 807 Interviews in London (of 4,525) 3.1 Average PwC Interviews in London www.pwc.com /  HQ: New York, NY 456 Interviews in London (of 7,390) 3.1 Average Accenture Interviews in London www.accenture.com /  HQ: Dublin 437 Interviews in London (of 10,136) 3.0 Average ## Interview Questions in London Sort: RelevancePopular Date 27 Apr 2011 29 Dec 2012 We consider numbers from 1 to 1 million. How many digits 2 are there??10 Answers600 000940951654321: The total number of digits 1-999,999 is 5888889 Total 1-9 = 9 10-99 = 2*9*10 and so on so number of 2s is 5888889/9Show more responses654321: The total number of digits 1-999,999 is 5888889 Total 1-9 = 9 10-99 = 2*9*10 and so on so number of 2s is 5888889/9Each time you fix a digit 2 (in units, tens...), then you change the other 5 digits. So you'll have 100 000 number. Having that for the six placements, the result would be 600 000.600,000 6C6 with 6 digits, 9(6C5) with 5, (9^2)(6C4) with 4...etc. Add them up 600,000.11111110^6-9^6observe the recurrence relation. 1-9 there is 1 from 1-99 there is 10 1-9 intervals plus 10 digits from 2x. and then it continuous in exactly the same fashion add one 0 you get 10 times as many plus 10^(n-1) from the nth digit you add. (((((10+10)*10+100)*10+1,000)*10+10,000)*10+100,000)=600,000Consider from 0000000 to 9999999, the number of 0,1,2,3,4,5,6,7,8,9 present must be the same since they are all symmetric. prove: the position can be filled with 0 can be filled with 1, so the number of 0=the number of 1; the number of 1=the number of 2 for the same reason. ... so the number of 0-9 are all the same. There are 6*1million positions in total, thus 6 million positions, divided by 10, is 600,000. 2 Jan 2011 You roll two die: What is the probability of rolling a 10 and an 11 before rolling a 7?8 Answers17/132How did you get that?Individually 10 comes up 3/11 (4+6, 5+5, 6+4), 11 comes up 2/11, 7 comes up 6/11. Consider 10 first: Then only interested in 7,11: Probability 11 is 2/8, 7 is 6/8. Consider 11 first: Then only interested in 7,10: Probability 10 is 3/9, 7 is 6/9. Consider 7 first: Busted. So add 10 then 11, and 11 then 10: (3/11)*(2/8)+(2/11)*(3/99)=17/132.Show more responses5/11 let's say d is the probability that 10 or 11 comes up before 7. probability of rolling 10: (6+4,4+6,5+5) = 3/36 probability of rolling 11: (6+5,5+6): 2/36 probability of rolling 7: (6+1,1+6,5+2,2+5,4+3,3+4): 6/36 d = you roll 10 or you roll 11 or (you roll something other than 10,11,7 and roll 10 or 11 before 7) d = (3/36) + (2/36) + (25/36)*d (11/36)*d = (5/36) d= 5/11Anonbyfar is correct, 5/11 is the right answer but for the wrong questiondoesnt the answer by 'Anonbyfar' (17/132) give you the probability of rolling a 10 and 11 within two rolls? Shouldnt this be then related to the probability of rolling a 7 (6/11). I.e 17/72?P(10 or 11 before 7) = P(10 or 11 | 10 or 11 or 7) = [P(10) + P(11)]/[P(10)+P(11)+P(7)] P(10) = 3/36 4,6 5,5 6,4 P(11)= 2/36 5,6 6,5 P(7) = 6/36 1,6 2,5, 3,4 4,3 5,2 6,1 Answer: 5/11roll 10 or 11: 5 cases roll 7: 6 cases roll others: 25 in total: 36 cases So Pr = 5/36 + 25/36 *Pr Pr = 5/11 6 Oct 2013 If I write down all of the numbers from 1 to 1,000,000 on a page, how many times do I write down the digit 2?9 Answers(base 10 )log(x)/10*xI got 468599600000 (verified through brute force)Show more responsesimagine the numbers written as 000001 000002 000003.....999998 999999, then we 1 million numbers containing 6 digits, each digit of 10 digits we have, is used as much as any other number. Therefore, you write down the digit 2 10% * 6 million - 600,000the minus must be an equal sign on the end, so 600,000 is the right number[1 x 10 x 10 x 10 x 10 x 10] + [10 x 1 x 10 x 10 x 10 x 10] - 1 + (bcos 222222 covered above) [10 x 10 x 1 x 10 x 10 x 10] - 2 + (bcos 222222 & 22222 covered) [10 x 10 x 10 x 1 x 10 x 10] - 3 [10 x 10 x 10 x 10 x 1 x 10] - 4 [10 x 10 x 10 x 10 x 10 x 1] - 5 = 600000 - 15 = 599985Continuation from above from 1 upto 1 with n 'zeros' (e.g. 10 has 1 zero, 100 has 2 zero, etc), answer = n(10^[n-1]) - 0.5n(n-1) So upto 1000000 would give 6 x 10 ^ 5 - 0.5 x 6 x 5 = 599985The idea is to exploit the symmetry among 0,1,2...9. Instead of 1->10^6, one should think of 000000 -> 999999, where 0-9 occurs with the same probablity . Altogether 000000 -> 999999 has 10^6 numbers, 6*10^6 digits; divided by 10 this is 600,000.The idea is recursion, let x_n be number of times that digit 2 appear in 0 to the largest n digit number. For example, x_1 = number of times that 2 appears in 0-9. x_2 is up until 99, etc. you work out to see x_1 = 1. x_2 = 10*x_1 + 10 etc. finally, I think you will get 600000 ### Assista at Jane Street was asked... 14 May 2012 2nd Round: You have a deck of cards, 26 red, 26 black. These are turned over, and at any point you may stop and exclaim "The next card is red.". If the next card is red you win £10. What's the optimal strategy? Prove this is the optimal strategy.6 AnswersThe simplest strategy is not to wait until any card is open and say the next one is red, than you have 26/52 chance of success. All other strategies require conditional probabilities, so when multiplying by the probability of that condition the overall probability is not going to be greater than 1/2.Regardless of your strategy, your probability of winning is 1/2Nope the optimal strategy is to count the amount of black cards that come up until you hit 26 and obviously the next one is redShow more responsesAdrian is wrong, because the strategy only work 50% of the time, when the last card is not black. So you still get 1/2.The idea here is A LEAP OF FAITH: that is we have found the optimal strategy when there is X red and Y black left in the deck. let Strategy(X,Y) be that strategy: it is either "Wait" or "Say it now". Let P(X,Y) be the probability that we are right by following this strategy. now Strategy(X, 0) is definitely "say it now" and P(X, 0) = 1. what about Strategy(1, 1)? half the time it will show black and we win for sure; half the time it will show red and we are doomed. Both action, "wait" and "say it now" have the same expected chance of winning, which is 1/2. so P(1,1) = 1/2. What about Strategy(2, 1)? if it is "wait", 2/3 time we go to (1, 1) and 1/3 time we go (2, 0). the chance of winning if we wait is 2/3*1/2 + 1/3*1 = 2/3. if it "say it now", the chance of winning is 2/3 as well. so P(2, 1) = 2/3. As you can see, we basically have the following relationships: P(X, 0) =1 for any X. P(X, Y) = max( X/(X+Y), X/(X+Y) * P(X-1,Y) + Y/(X+Y) * P(X, Y-1)) and you can probably guess that P(X, Y) = X/(X+Y) so P(26, 26) = 1/2. That means calling out on first card or wait both gives you optimal strategy.always 1/2 ### Analyst at J.P. Morgan was asked... 3 Apr 2011 puzzle 9 balls, identical in appearance. 8 are of same weight. 1 is heavier. Identify this heavier ball by using a scale twice.5 Answersput 9 balls in groups of 3.Split them into two sets of 4 and 1 individual ball. Place the two sets on each side of the balance. If the sets balance out, then the individual ball is the heavier one. If the sets don't balance, then take the heavier set and split it up into two sets of two and repeat.I believe that an explanation to the solution that Interview Candidate is suggesting can be found here: http://brainteaserbible.com/snooker-balls-weighing-scalesShow more responsesSplit the balls into groups of three, compare two of the groups. If the sets balance then the heavy ball is in the third group, otherwise it is in the group left out. Whichever it is, take two balls from that group to weigh. If they balance, you know it is the third ball, if they don't, you have the heavier ball.http://brainteaserbible.com/snooker-balls-weighing-scales ### Financial Software Developer at Bloomberg L.P. was asked... 4 Dec 2009 write a function that returns the first unique element in an array5 AnswersWhat the type of elements in the array? If character, set up the hash table, and scan the array. the hash table stores the index of each element. If the element appears more than once, update the table as a negetive value. After scanning, find the smallest index value from the hash table, which would be the first uniqure element. The time complex gonna be O(n), where n is the length of array.I wrote a sample program here. (I use map container here to replace hash_map. The doesnot work on my computer) #include "stdafx.h" #include #include using namespace std; static int find_unique_ele(char*,int); int _tmain(int argc, _TCHAR* argv[]) { char test[6] = {'a','b','c','b','c','a'}; int min_index = find_unique_ele(test,6); if (min_index store_table_value; map store_table; while(i ::const_iterator test_it = store_table.find(*test); if(test_it != store_table.end()){ store_table[*test] = -1; } else{ store_table.insert(map::value_type(store_table_value(*test,i))); } i++; test++; } map::const_iterator table_it = store_table.begin(); store_table_value temp_value = *table_it; int min_index = temp_value.second; while(++table_it != store_table.end()){ temp_value = *table_it; if(temp_value.second < min_index || min_index < 0){ min_index = temp_value.second; } } return min_index; }I think in the above solution you have to scan the original array twice making it order (2n) or O(n). The 2nd scan is needed to determine the corresponding entry in the hash table whether it is unique or not. We need this to determine the first unique element in the array.Show more responsesSort the list then scan the list looking for a data item that does not match the previous item or the next item.very easy with c++ stl using unique() function or use hashtable for O(1) look -up HashTable mm = new HashTable(myArray.count()); for(int i =0; i 25 Apr 2013 The questions were not very difficult but you really need to have all the concepts crystal-clear and be ready to apply them successfully. One of the questions was "how to count the letters in this string:" "The quick brown fox jumps over the lazy dog";11 Answerspublic static int countWords(String str){ if(str == null || str.isEmpty()) return 0; int count = 0; for(int e = 0; e < str.length(); e++){ if(str.charAt(e) != ' '){ count++; while(str.charAt(e) != ' ' && e < str.length()-1){ e++; } }else{ e++; } } return count; }Sorry, the above version has an error!!!!!!!!!!!!!!!!!!!!!!!! concider this one: public static int countWords(String str){ if(str == null || str.isEmpty()) return 0; int count = 0; for(int e = 0; e < str.length(); e++){ if(str.charAt(e) != ' '){ count++; while(str.charAt(e) != ' ' && e < str.length()-1){ e++; } } } return count; }# That's why i love Python: len(re.findall('[a-zA-Z]', s))Show more responsessorry u need to give input ;) ------ len(re.findall('[a-zA-Z]', "The quick brown fox jumps over the lazy dog"))It depends on what has to be considered a word. For example, if we consider a word any string between spaces, we can write it in a more compact way: public static int countWords(String str){ if(str == null || str.isEmpty()) return 0; return str.split(" ").length; }The question is how to count the characters in the string. If we can assume that the input is ASCII - always clarify first - then we know that the character range is 0-255. void count(char* str, int counts[255]) { if (str == 0) return; if (counts == 0) return; for (char *c = str; c != 0; c++) { counts[*c]++; } } First we assert the input is valid. Note that we take an additional parameter - an array of the count of each ASCII character. We loop through the string until we reach the null terminator, and we iterate a char pointer through each character in the string. For each iteration, we increment the counter in the array. Memory complexity is O(1), runtime complexity is O(n). If the input must be unicode, then we may consider alternatively using a hash table. void count(wchar_t* str, std::unordered_map& counts) { if (str == 0) return; for (char *c = str; c != 0; c++) { counts[*c]++; } } We still arrive at an O(1) memory complexity and O(n) runtime complexity (although it is worth noting that despite a hash table lookup takes O(1) like an array, the fixed cost of each lookup is higher for the hash table due to the hashing function, and in the worst case a lookup can be O(n)).The question asks to count the characters, without using String.length() you can do this: public static Pair countLetters(String s) { //If the string is null or it is empty then it will have no character if (s == null || s.isEmpty()) { //So return the pair with 0 and 0 return new Pair(0, 0); } //If we should still run the loop. boolean run = true; //The count of characters int count = 0; //The count of characters without spaces int countWithoutSpace = 0; //While we are still running (we are at a valid index) while (run) { //Then try to try { //Get the character at the current count char c = s.charAt(count); //Add one to the count as we have a valid character count++; //If the character is not a space if (c != ' ') { //Then add one to the count without spaces. countWithoutSpace++; } //If we get a StringIndexOutOfBooundsException it means that the current count is outside the length of the string. } catch (StringIndexOutOfBoundsException e) { //So stop the loop. run = false; } } //And return the pair with the count and the count without spaces. return new Pair(count, countWithoutSpace); } This returns both the count with and without spaces and does not use a built in length function.Wait..Am I missing something? string.Length will do the job?Yeah i think string.length() will do the job.failed! String.length returns the length of the String. Google asked you how many letters - in other words you cannot count the spaces. String. length returns number of unicode characters -and so includes spacesI would do this: public static int countWords(String sentence){ String noSpace; //REMOVE SPACE noSpace = sentence.replaceAll(" ", ""); return noSpace.length }

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# Fraction Math Chat Cards (Determining Parts of the Whole and Improper Fractions) Subjects Grade Levels Resource Types Common Core Standards Product Rating Not yet rated File Type Presentation (Powerpoint) File Be sure that you have an application to open this file type before downloading and/or purchasing. 4.9 MB   |   69 pages ### PRODUCT DESCRIPTION This Fraction Chat Card presentation offers slides that will prompt discussion on two of the most important Big Understandings in Fractions: 1) The parts of a whole (for example, sixths) must be equivalent. 2) When using symbolic notation to represent fractions the top number (numerator) represents how many you have and the bottom number (denominator) represents what kind of fractional part you have. Without a solid understanding of these concepts, it is nearly impossible for students to move on to higher level thinking when it comes to fractions. These Chat Cards can be used as your mini-lesson over a couple of days, as review for previously taught concepts and as a jumping off point for Math Chats. The Cards can also be helpful for a student that missed a day of class or families that want to learn more about the concepts being taught in class. More Fraction Chat Cards will be coming soon, covering Big Ideas such as equivalency vs. congruency, determining equivalent fractions, comparing and ordering fractions, and more! Total Pages 69 Answer Key Does Not Apply Teaching Duration N/A ### Average Ratings N/A Overall Quality: N/A Accuracy: N/A Practicality: N/A Thoroughness: N/A Creativity: N/A Clarity: N/A Total: 0 ratings COMMENTS AND RATINGS: Please log in to post a question. PRODUCT QUESTIONS AND ANSWERS: \$3.25 Digital Download User Rating: 4.0/4.0 (14 Followers) \$3.25 Digital Download Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. Learn More Sign up

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Search by Topic Resources tagged with Area similar to Once Upon a Time: Filter by: Content type: Stage: Challenge level: There are 93 results Broad Topics > Measures and Mensuration > Area Tiling Stage: 2 Challenge Level: An investigation that gives you the opportunity to make and justify predictions. Area and Perimeter Stage: 2 Challenge Level: What can you say about these shapes? This problem challenges you to create shapes with different areas and perimeters. Fitted Stage: 2 Challenge Level: Nine squares with side lengths 1, 4, 7, 8, 9, 10, 14, 15, and 18 cm can be fitted together to form a rectangle. What are the dimensions of the rectangle? Uncanny Triangles Stage: 2 Challenge Level: Can you help the children find the two triangles which have the lengths of two sides numerically equal to their areas? Torn Shapes Stage: 2 Challenge Level: These rectangles have been torn. How many squares did each one have inside it before it was ripped? Tiles on a Patio Stage: 2 Challenge Level: How many ways can you find of tiling the square patio, using square tiles of different sizes? Extending Great Squares Stage: 2 and 3 Challenge Level: Explore one of these five pictures. My New Patio Stage: 2 Challenge Level: What is the smallest number of tiles needed to tile this patio? Can you investigate patios of different sizes? Blue and White Stage: 3 Challenge Level: Identical squares of side one unit contain some circles shaded blue. In which of the four examples is the shaded area greatest? Tiling Into Slanted Rectangles Stage: 2 and 3 Challenge Level: A follow-up activity to Tiles in the Garden. Ribbon Squares Stage: 2 Challenge Level: What is the largest 'ribbon square' you can make? And the smallest? How many different squares can you make altogether? Rati-o Stage: 3 Challenge Level: Points P, Q, R and S each divide the sides AB, BC, CD and DA respectively in the ratio of 2 : 1. Join the points. What is the area of the parallelogram PQRS in relation to the original rectangle? Pebbles Stage: 2 and 3 Challenge Level: Place four pebbles on the sand in the form of a square. Keep adding as few pebbles as necessary to double the area. How many extra pebbles are added each time? Numerically Equal Stage: 2 Challenge Level: Can you draw a square in which the perimeter is numerically equal to the area? A Square in a Circle Stage: 2 Challenge Level: What shape has Harry drawn on this clock face? Can you find its area? What is the largest number of square tiles that could cover this area? Poly-puzzle Stage: 3 Challenge Level: This rectangle is cut into five pieces which fit exactly into a triangular outline and also into a square outline where the triangle, the rectangle and the square have equal areas. Lawn Border Stage: 1 and 2 Challenge Level: If I use 12 green tiles to represent my lawn, how many different ways could I arrange them? How many border tiles would I need each time? More Transformations on a Pegboard Stage: 2 Challenge Level: Use the interactivity to find all the different right-angled triangles you can make by just moving one corner of the starting triangle. Dicey Perimeter, Dicey Area Stage: 2 Challenge Level: In this game for two players, you throw two dice and find the product. How many shapes can you draw on the grid which have that area or perimeter? Through the Window Stage: 2 Challenge Level: My local DIY shop calculates the price of its windows according to the area of glass and the length of frame used. Can you work out how they arrived at these prices? Cover the Tray Stage: 2 Challenge Level: These practical challenges are all about making a 'tray' and covering it with paper. Tiles in the Garden Stage: 2 Challenge Level: How many tiles do we need to tile these patios? How Random! Stage: 2 Challenge Level: Explore this interactivity and see if you can work out what it does. Could you use it to estimate the area of a shape? Isosceles Triangles Stage: 3 Challenge Level: Draw some isosceles triangles with an area of $9$cm$^2$ and a vertex at (20,20). If all the vertices must have whole number coordinates, how many is it possible to draw? Framed Stage: 3 Challenge Level: Seven small rectangular pictures have one inch wide frames. The frames are removed and the pictures are fitted together like a jigsaw to make a rectangle of length 12 inches. Find the dimensions of. . . . Fence It Stage: 3 Challenge Level: If you have only 40 metres of fencing available, what is the maximum area of land you can fence off? Rope Mat Stage: 2 Challenge Level: How many centimetres of rope will I need to make another mat just like the one I have here? Geoboards Stage: 2 Challenge Level: This practical challenge invites you to investigate the different squares you can make on a square geoboard or pegboard. F'arc'tion Stage: 3 Challenge Level: At the corner of the cube circular arcs are drawn and the area enclosed shaded. What fraction of the surface area of the cube is shaded? Try working out the answer without recourse to pencil and. . . . Wrapping Presents Stage: 2 Challenge Level: Choose a box and work out the smallest rectangle of paper needed to wrap it so that it is completely covered. Making Boxes Stage: 2 Challenge Level: Cut differently-sized square corners from a square piece of paper to make boxes without lids. Do they all have the same volume? Bull's Eye Stage: 3 Challenge Level: What fractions of the largest circle are the two shaded regions? Making Squares Stage: 2 Investigate all the different squares you can make on this 5 by 5 grid by making your starting side go from the bottom left hand point. Can you find out the areas of all these squares? Dissect Stage: 3 Challenge Level: It is possible to dissect any square into smaller squares. What is the minimum number of squares a 13 by 13 square can be dissected into? Fencing Lambs Stage: 2 Challenge Level: A thoughtful shepherd used bales of straw to protect the area around his lambs. Explore how you can arrange the bales. Being Curious - Primary Measures Stage: 1 and 2 Challenge Level: Measure problems for inquiring primary learners. Triangle Island Stage: 2 Challenge Level: You have pitched your tent (the red triangle) on an island. Can you move it to the position shown by the purple triangle making sure you obey the rules? Warmsnug Double Glazing Stage: 3 Challenge Level: How have "Warmsnug" arrived at the prices shown on their windows? Which window has been given an incorrect price? Cutting it Out Stage: 1 and 2 Challenge Level: I cut this square into two different shapes. What can you say about the relationship between them? Circle Panes Stage: 2 Challenge Level: Look at the mathematics that is all around us - this circular window is a wonderful example. Being Resilient - Primary Measures Stage: 1 and 2 Challenge Level: Measure problems at primary level that may require resilience. Square Areas Stage: 3 Challenge Level: Can you work out the area of the inner square and give an explanation of how you did it? Triangle Relations Stage: 2 Challenge Level: What do these two triangles have in common? How are they related? A Day with Grandpa Stage: 2 Challenge Level: Grandpa was measuring a rug using yards, feet and inches. Can you help William to work out its area? Maths Filler Stage: 3 Challenge Level: Imagine different shaped vessels being filled. Can you work out what the graphs of the water level should look like? Being Resourceful - Primary Measures Stage: 1 and 2 Challenge Level: Measure problems at primary level that require careful consideration. Being Collaborative - Primary Measures Stage: 1 and 2 Challenge Level: Measure problems for primary learners to work on with others. Changing Areas, Changing Perimeters Stage: 3 Challenge Level: How can you change the area of a shape but keep its perimeter the same? How can you change the perimeter but keep the area the same? Different Sizes Stage: 1 and 2 Challenge Level: A simple visual exploration into halving and doubling. Perimeter Possibilities Stage: 3 Challenge Level: I'm thinking of a rectangle with an area of 24. What could its perimeter be?

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