Mark Zegarelli

Mark Zegarelli is a math teacher and tutor with degrees in math and English from Rutgers University. He is the author of a dozen books, including Basic Math & Pre-Algebra For Dummies, SAT Math For Dummies, and Calculus II For Dummies. Through online tutoring, he teaches multiplication and beyond to preschoolers in a way that sets them up for school success while keeping the natural magic of math alive. Contact Mark at markzegarelli.com.

Articles From Mark Zegarelli

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269 results
269 results
Digital SAT Math Prep For Dummies Cheat Sheet

Cheat Sheet / Updated 10-05-2023

Although there's no shortcut to success on the math sections of the SAT, you can study and prepare in order to get the best SAT score you possibly can. Knowing what will be on the test (and what won't be) is key so you know what to brush up on. Also, some basic strategy goes a long way toward helping you get the best score you can. Finally, mapping out a time-management plan to answer (and skip!) the right questions can really boost your score.

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Pre-Algebra Practice Questions: Finding the Volume of Prisms and Cylinders

Article / Updated 08-07-2023

To find the volume of a prism or cylinder, you can use the following formula, where Ab is the area of the base and h is the height: V = Ab x h Practice questions Find the volume of a prism with a base that has an area of 6 square centimeters and a height of 3 centimeters. Figure out the approximate volume of a cylinder whose base has a radius of 7 millimeters and whose height is 16 millimeters. Answers and explanations 18 cubic centimeters V = Ab x h = 6cm2 x 3cm = 18cm3 Approximately 2,461.76 cubic millimeters First, use the area formula for a circle to find the area of the base: Ab = π x r2 ≅ 3.14 x (7mm)2 = 3.14 x 49mm2 = 153.86mm2 Plug this result into the formula for the volume of a cylinder: V = Ab x h = 153.86mm2 x 16mm = 2,461.76mm3

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Finding the Area of a Surface of Revolution

Video / Updated 07-14-2023

The nice thing about finding the area of a surface of revolution is that there’s a formula you can use. Memorize it and you’re halfway done. To find the area of a surface of revolution between a and b, watch this video tutorial or follow the steps below: This formula looks long and complicated, but it makes more sense when you spend a minute thinking about it. The integral is made from two pieces: The arc-length formula, which measures the length along the surface The formula for the circumference of a circle, which measures the length around the surface So multiplying these two pieces together is similar to multiplying length and width to find the area of a rectangle. In effect, the formula allows you to measure surface area as an infinite number of little rectangles. When you’re measuring the surface of revolution of a function f(x) around the x-axis, substitute r = f(x) into the formula: For example, suppose that you want to find the area of revolution that’s shown in this figure. Measuring the surface of revolution of y = x3 between x = 0 and x = 1. To solve this problem, first note that for So set up the problem as follows: To start off, simplify the problem a bit: You can solve this problem by using the following variable substitution: Now substitute u for 1+ 9x4 and for x3 dx into the equation: Notice that you change the limits of integration: When x = 0, u = 1. And when x = 1, u = 10. Now you can perform the integration: Finally, evaluate the definite integral:

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Evaluating an Expression with Only Multiplication & Division

Article / Updated 07-10-2023

Some expressions contain only multiplication and division. When this is the case, the rule for evaluating the expression is pretty straightforward. When an expression contains only multiplication and division, evaluate it step by step from left to right. The Three Types of Big Four Expressions Expression Example Rule Contains only addition and subtraction 12 + 7 – 6 – 3 + 8 Evaluate left to right. Contains only multiplication and division 18 ÷ 3 x 7 ÷ 14 Evaluate left to right. Mixed-operator expression: contains a combination of addition/subtraction and multiplication/division 9 + 6 ÷ 3 1. Evaluate multiplication and division left to right. 2. Evaluate addition and subtraction left to right. Suppose you want to evaluate this expression: 9 × 2 ÷ 6 ÷ 3 × 2 Again, the expression contains only multiplication and division, so you can move from left to right, starting with 9 x 2: = 18 ÷ 6 ÷ 3 × 2 = 3 ÷ 3 × 2 = 1 × 2 = 2 Notice that the expression shrinks one number at a time until all that’s left is 2. So 9 × 2 ÷ 6 ÷ 3 × 2 = 2 Here’s another quick example: −2 × 6 ÷ −4 Even though this expression has some negative numbers, the only operations it contains are multiplication and division. So you can evaluate it in two steps from left to right (remembering the rules for multiplying and dividing with negative numbers): = −2 × 6 ÷ −4 = −12 ÷ −4 = 3 Thus, −2 × 6 ÷ −4 = 3

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Pre-Algebra: Comparing Fractions Using Cross-Multiplication

Article / Updated 07-10-2023

Cross-multiplication is a handy tool for finding the common denominator for two fractions, which is important for many operations involving fractions. In the following practice questions, you are asked to cross-multiply to compare fractions to find out which is greater or less. Practice questions 1. Find the lesser fraction: 2. Among these three fractions, which is greatest: Answers and explanations 1. Of the two fractions, Cross-multiply to compare the two fractions: Because 35 is less than 36, 2. Of the three fractions, Use cross-multiplication to compare the first two fractions. Because 21 is greater than 20, this means that 1/10 is greater than 2/21, so you can rule out 2/21. Next, compare 1/10 and 3/29 by cross-multiplying. Because 30 is greater than 29, 3/29 is greater than 1/10. Therefore, 3/29 is the greatest of the three fractions.

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Important Operations that Make Math Problems Easier

Article / Updated 07-10-2023

The Big Four math operations — addition, subtraction, multiplication, and division — let you combine numbers and perform calculations. Certain operations possess properties that enable you to manipulate the numbers in the problem, which comes in handy, especially when you get into higher math like algebra. The important properties you need to know are the commutative property, the associative property, and the distributive property. Understanding what an inverse operation is is also helpful. Inverse operations Inverse operations are pairs of operations that you can work "backward" to cancel each other out. Two pairs of the Big Four operations — addition, subtraction, multiplication, and division —are inverses of each other: Addition and subtraction are inverse operations of each other. When you start with any value, then add a number to it and subtract the same number from the result, the value you started with remains unchanged. For example: 2 + 3 = 5 so 5 – 3 = 2 7 – 1 = 6 so 6 + 1 = 7 Multiplication and division are inverse operations of each other. When you start with any value, then multiply it by a number and divide the result by the same number (except zero), the value you started with remains unchanged. For example: 3 × 4 = 12 so 12 ÷ 4 = 3 10 ÷ 2 = 5 so 5 × 2 = 10 The commutative property An operation is commutative when you apply it to a pair of numbers either forwards or backwards and expect the same result. The two Big Four that are commutative are addition and subtraction. Addition is commutative because, for example, 3 + 5 is the same as 5 + 3. In other words 3 + 5 = 5 + 3 Multiplication is commutative because 2 × 7 is the same as 7 × 2. In other words 2 × 7 = 7 × 2 The associative property An operation is associative when you can apply it, using parentheses, in different groupings of numbers and still expect the same result. The two Big Four operations that are associative are addition and multiplication. Addition is associative because, for example, the problem (2 + 4) + 7 produces the same result as does the problem 2 + (4 + 7). In other words, (2 + 4) + 7 = 2 + (4 + 7) No matter which pair of numbers you add together first, the answer is the same: 13. Multiplication is associative because, for example, the problem 3 × (4 × 5) produces the same result as the problem (3 × 4) × 5. In other words, 3 × (4 × 5) = (3 × 4) × 5 Again, no matter which pair of numbers you multiply first, both problems yield the same answer: 60. The distributive property The distributive property connects the operations of multiplication and addition. When multiplication is described as "distributive over addition," you can split a multiplication problem into two smaller problems and then add the results. For example, suppose you want to multiply 27 × 6. You know that 27 equals 20 + 7, so you can do this multiplication in two steps: First multiply 20 × 6; then multiply 7 × 6. 20 × 6 = 1207 × 6 = 42 Then add the results. 120 + 42 = 162 Therefore, 27 × 6 = 162.

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Solving Simple Equations in Pre-Algebra Problems

Article / Updated 06-28-2023

When dealing with simple algebraic expressions, you don't always need algebra to solve them. The following practice questions ask you to use three different methods: inspecting, rewriting the problem, and guessing and checking. Practice questions In the following questions, solve for x in each case just by looking at the equation. 1. 18 – x = 12 2. 4x = 44 In the following questions, use the correct inverse operation to rewrite and solve each problem. 3. 100 – x = 58 4. 238/x = 17 In the following questions, find the value of x in each equation by guessing and checking. 5. 12x – 17 = 151 6. 19x – 8 = 600 Answers and explanations x = 6You can solve this problem through simple inspection. Because 18 – 6 = 12, x = 6. x = 11Again, through simple inspection, because 4(11) = 44, you know that x = 11. x = 42Turn the problem around by changing the subtraction to addition: 100 – x = 58 means the same thing as 100 – 58 = x, so x = 42. x = 14Turn the problem around by switching around the division: so x = 14. x = 14Guess what you think the answer may be. For example, perhaps it's x = 10: 12(10) – 17 = 120 – 17 = 103 103 is less than 151, so this guess is too low. Try x = 20: 12(20) – 17 = 240 – 17 = 223 223 is greater than 151, so this guess is too high. Therefore, x is between 10 and 20. Try x = 15: 12(15) – 17 = 180 – 17 = 163 163 is a little greater than 151, so this guess is a little too high. Try x = 14: 12(14) – 17 = 168 – 17 = 151 151 is correct, so x = 14. x = 32Again, start by guessing. First, try x = 10: 19(10) – 8 = 190 – 8 = 182 182 is much less than 600, so this guess is much too low. Try x = 30: 19(30) – 8 = 570 – 8 = 562 562 is still less than 600, so this guess is still too low. Try x = 35: 19(35) – 8 = 665 – 8 = 657 657 is greater than 600, so this guess is too high. Therefore, x is between 30 and 35. Try x = 32: 19(32) – 8 = 608 – 8 = 600 600 is correct, so x = 32.

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Applying Order of Operations in Expressions with Exponents

Article / Updated 06-22-2023

Evaluate exponents from left to right before you begin evaluating Big Four operations (adding, subtracting, multiplying, and dividing). The trick here is to turn the expression into a Big Four expression and then use the other basic order of operations rules. The Three Types of Big Four Expressions Expression Example Rule Contains only addition and subtraction 12 + 7 - 6 - 3 + 8 Evaluate left to right. Contains only multiplication and division 18 ÷ 3 x 7 ÷ 14 Evaluate left to right. Mixed-operator expression: contains a combination of addition/subtraction and multiplication/division 9 + 6 ÷ 3 1. Evaluate multiplication and division left to right. 2. Evaluate addition and subtraction left to right. For example, suppose you want to evaluate the following: 3 + 5² - 6 First, evaluate the exponent: 3 + 25 - 6 At this point, the expression contains only addition and subtraction, so you can evaluate it from left to right in two steps: = 28 - 6 = 22 So 3 + 5² - 6 = 22

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Rearranging Algebraic Equations to Isolate X

Article / Updated 06-20-2023

A quick method for solving algebra problems is to re-arrange the equation by placing all x terms on one side of the equal sign and all constants (non-x terms) on the other side. Essentially, you're doing the addition and subtraction without showing it. You can then isolate x. Practice questions Rearrange the equation 10x + 5 = 3x + 19 to solve for x. Solve –[2(x + 7) + 1] = x – 12 for x. Answers and explanations x = 2First, rearrange the terms of the equation so that the x terms are on one side and the constants are on the other. In this case, you can do this in two steps: Second, combine like terms on both sides: 7x = 14 The third and final step is to divide (in this case, by 7) to isolate x: x = –1Before you can begin rearranging terms, remove the parentheses on the left side of the equation. Start with the inner parentheses, multiplying 2 by every term inside that set: –[2(x + 7) + 1] = x – 12 –[2x + 14 + 1] = x – 12 Next, remove the remaining parentheses, switching the sign of every term within that set: –2x – 14 – 1 = x – 12 Now you can solve for x by, first, rearranging the terms of the equation: –2x – 14 – 1 + 12 = x –14 – 1 + 12 = x + 2x Then you combine like terms on both sides: –3 = 3x Finally, you divide (in this example, by 3) to isolate x:

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Pre-Algebra: Cross-Multiply to Solve Equations

Article / Updated 06-06-2023

When an algebraic equation contains fractions, you can use cross-multiplication to solve the equation. The following practice questions contain two equal fractions, where you need to cross-multiply to solve them. Practice questions Rearrange the equation to solve for x. Solve the equation Answers and explanations x = –4Remove the fraction from the equation by cross-multiplying: Multiply to remove the parentheses from the left side of the equation: 8x + 40 = –2x At this point, you can solve for x: x = 5Remove the fractions from the equation by cross-multiplying: Remove the parentheses from the left side of the equation by multiplying through by x; remove parentheses from the right side of the equation by FOILing: 4x2 – 7x = 4x2 – 10x + 6x – 15 Rearrange the equation: 4x2 – 7x – 4x2 + 10x – 6x = –15 Notice that the two x2 terms cancel each other out:

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