SAT Math : Variables

Example Questions

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Example Question #1 : Variables

Multiply:

Explanation:

Distribute the monomial through the polynomial.

Example Question #2 : Variables

Evaluate:

Explanation:

Distribute the monomial through each term inside the parentheses.

Example Question #1 : Monomials

Multiply the monomials:

Explanation:

In order to multiply this, we can simply multiply the coefficients together and the  variables together.  Anytime we multiply a variable of the same base, we can add the exponents.

Simplify the right side.

Example Question #3 : Variables

Distribute:

Explanation:

Distribute the monomial to each part of the polynomial, paying careful attention to signs:

Example Question #4 : Variables

Find the product:

Explanation:

Find the product:

Step 1: Multiply the numerators and denominators using the properties of exponents. (When multiplying exponents, add them.)

Step 2: Simplify the expression.

Example Question #1 : How To Use The Direct Variation Formula

Phillip can paint  square feet of wall per minute. What area of the wall can he paint in 2.5 hours?

Explanation:

Every minute Phillip completes another  square feet of painting. To solve for the total area that he completes, we need to find the number of minutes that he works.

There are 60 minutes in an hour, and he paints for 2.5 hours. Multiply to find the total number of minutes.

If he completes square feet per minute, then we can multiply by the total minutes to find the final answer.

Example Question #1 : Variables

The value of  varies directly with the square of and the cube of . If  when and , then what is the value of  when  and ?

Explanation:

Let's consider the general case when y varies directly with x. If y varies directly with x, then we can express their relationship to one another using the following formula:

y = kx, where k is a constant.

Therefore, if y varies directly as the square of x and the cube of z, we can write the following analagous equation:

y = kx2z3, where k is a constant.

The problem states that y = 24 when x = 1 and z = 2. We can use this information to solve for k by substituting the known values for y, x, and z.

24 = k(1)2(2)3 = k(1)(8) = 8k

24 = 8k

Divide both sides by 8.

3 = k

k = 3

Now that we have k, we can find y if we know x and z. The problem asks us to find y when x = 3 and z = 1. We will use our formula for direct variation again, this time substitute values for k, x, and z.

y = kx2z3

y = 3(3)2(1)3 = 3(9)(1) = 27

y = 27

Example Question #1 : How To Use The Direct Variation Formula

In a growth period, a population of flies triples every week. If the original population had 3 flies, how big is the population after 4 weeks?

Explanation:

We know that the initial population is 3, and that every week the population will triple.

The equation to model this growth will be , where is the initial size, is the rate of growth, and is the time.

In this case, the equation will be .

Alternatively, you can evaluate for each consecutive week.

Week 1:

Week 2:

Week 3:

Week 4:

Example Question #2 : How To Use The Direct Variation Formula

and are the diameter and circumference, respectively, of the same circle.

Which of the following is a true statement? (Assume all quantities are positive)

varies directly as the fourth root of .

varies directly as the fourth power of .

varies inversely as the fourth power of .

varies directly as .

varies inversely as the fourth root of .

varies directly as .

Explanation:

If and are the diameter and circumference, respectively, of the same circle, then

.

By substitution,

Taking the square root of both sides:

Taking as the constant of variation, we get

,

meaning that varies directly as .

Example Question #1 : How To Use The Direct Variation Formula

is the radius of the base of a cone; is its height; is its volume.

.

Which of the following is a true statement?

varies directly as the third power of .

varies directly as the fifth power of .

varies directly as the fifth root of .

varies directly as the cube root of .

varies directly as .

varies directly as the fifth power of .

Explanation:

The volume of a cone can be calculated from the radius of its base , and the height , using the formula

, so .

, so .

, so by substitution,

Square both sides:

If we take  as the constant of variation, then

,

and varies directly as the fifth power of .

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