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SAT Math : Variables

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Example Questions

Example Question #1 : Variables

Multiply: (1-x^2)(9x)

Possible Answers:

-1-9x^3

9x-9x^3

-x^2+9x+1

9x^3-9x

9x^3-1

Correct answer:

9x-9x^3

Explanation:

Distribute the monomial through the polynomial.

(1-x^2)(9x)= 9x-9x^3

The answer is: 9x-9x^3

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Example Question #2 : Variables

Evaluate: 3x(x^2+3x-1)

Possible Answers:

3x^3+x^2-x

4x^3+6x-3x

3x^3+9x^2-3x

3x^3-9x^2+3x

4x^3-6x-3x

Correct answer:

3x^3+9x^2-3x

Explanation:

Distribute the monomial through each term inside the parentheses.

3x(x^2+3x-1)= 3x(x^2)+(3x)(3x)+(3x)(-1)

= 3x^3+9x^2-3x

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Example Question #2 : Variables

Multiply the monomials: $6x \times 4x^2$

Possible Answers:

$\textup{They are not like terms, so they can't be multiplied.}$

24x^3

48x^3

72x

24x^2

Correct answer:

24x^3

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.

$6x \times 4x^2 = (6\cdot 4) (x^1\cdot x^2)$

Simplify the right side.

$(6\cdot 4) (x^1\cdot x^2) = 24(x^{1+2})= 24x^3$

The answer is: 24x^3

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Example Question #1 : Variables

Distribute: 4ab(3a-5b+1)

Possible Answers:

-8ab-1

12a^2b-20ab^2

12a^2b+20ab^2+4ab

12a^2b-20ab^2+4ab

-8ab+1

Correct answer:

12a^2b-20ab^2+4ab

Explanation:

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

4ab(3a-5b+1)

$4ab\cdot3a+4ab\cdot(-5b)+4ab\cdot1$

12a^2b-20ab^2+4ab

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Example Question #2131 : Sat Mathematics

Find the product:

$\left ( \frac{5x^3y^5}{y^6z^4} \right )\left ( \frac{2xy^{-1}}{3zx^2} \right )$

Possible Answers:

x^2z^4

$\frac{10x^2}{3y^2z^5}$

$\frac{5y^6}{2xz^4}$

$\frac{10x^3}{z^4}$

$\frac{x}{y^4}$

Correct answer:

$\frac{10x^2}{3y^2z^5}$

Explanation:

Find the product:

$\left ( \frac{5x^3y^5}{y^6z^4} \right )\left ( \frac{2xy^{-1}}{3zx^2} \right )$

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

$\frac{5x^3y^5\cdot 2xy^{-1}}{y^6z^4\cdot 3zx^2}$

$\frac{10x^4y^4}{3y^6z^5x^2}$

Step 2: Simplify the expression.

$\frac{10x^2}{3y^2z^5}$

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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?

Possible Answers:

$25y\ ft^2$

$50y\ ft^2$

$300y\ ft^2$

$2.5y\ ft^2$

$150y\ ft^2$

Correct answer:

$150y\ ft^2$

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.

$(60\frac{min}{hr})(2.5hr)=150min$

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

$(y\frac{ft^2}{min})(150min)=150y\ ft^2$

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Example Question #1 : Direct And Inverse Variation

The value of varies directly with the square of and the cube of . If y=24 when x=1 and z=2 , then what is the value of when x=3 and z=1 ?

Possible Answers:

Correct answer:

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

The answer is 27.

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Example Question #2 : Direct And Inverse Variation

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?

Possible Answers:

$729$

$2187$

$243$

$81$

$27$

Correct answer:

$243$

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 i(r)^n , where is the initial size, is the rate of growth, and is the time.

In this case, the equation will be 3(3)^4 .

3(3)^4=3(81)=243

Alternatively, you can evaluate for each consecutive week.

Week 1: 3(3)=9

Week 2: 3(9)=27

Week 3: 3(27)=81

Week 4: 3(81)=243

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Example Question #8 : Variables

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

$d = 12L^{2}$

$C = t^{2}$

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

Possible Answers:

varies directly as .

varies directly as the fourth root of .

varies inversely as the fourth power of .

varies directly as the fourth power of .

varies inversely as the fourth root of .

Correct answer:

varies directly as .

Explanation:

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

$C = \pi d$ .

By substitution,

$t^{2} = \pi \cdot 12 L^{2}$

$t^{2} = 12 \pi L^{2}$

Taking the square root of both sides:

$\sqrt{t^{2} }= \sqrt{12 \pi L^{2}}$

$t = \sqrt{12 \pi} \cdot \sqrt{L^{2}}$

$t = \sqrt{12 \pi} \cdot L$

Taking $K= \sqrt{12 \pi}$ as the constant of variation, we get

t =K L ,

meaning that varies directly as .

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

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

$n = 4r = h^{2}$ ; $m = V ^{2}$ .

Which of the following is a true statement?

Possible Answers:

varies directly as the cube root of .

varies directly as the third power of .

varies directly as .

varies directly as the fifth root of .

varies directly as the fifth power of .

Correct answer:

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

$V = \frac{1}{3} \pi r^{2}h$

n = 4r , so $r = \frac{n}{4}$ .

$n = h^{2}$ , so $h = \sqrt{n}$ .

$V = \frac{1}{3} \pi r^{2}h$ , so by substitution,

$V = \frac{1}{3} \pi\left ( \frac{n}{4} \right ) ^{2} \cdot \sqrt{n}$

$V = \frac{1}{3} \pi\left ( \frac{n^{2}}{16} \right ) \cdot \sqrt{n}$

$V = \frac{1}{48} \pi \cdot n^{2}} \sqrt{n}$

Square both sides:

$V ^{2 }=\left ( \frac{1}{48} \pi \cdot n^{2}} \sqrt{n} \right )^{2}$

$m=\left ( \frac{1}{48} \pi \right )^{2} \cdot \left ( n^{2}} \sqrt{n} \right )^{2}$

$m=\left ( \frac{1}{48} \pi \right )^{2} \cdot \left ( n^{2}} \right )^{2} \cdot \left ( \sqrt{n} \right )^{2}$

$m=\left ( \frac{1}{48} \pi \right )^{2} \cdot n^{4}} \cdot n$

$m=\left ( \frac{1}{48} \pi \right )^{2} \cdot n^{5}}$

If we take $K = \left ( \frac{1}{48} \pi \right )^{2}$ as the constant of variation, then

$m=K n^{5}}$ ,

and varies directly as the fifth power of .

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SAT Math : Variables

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #1 : Variables

Example Question #2 : Variables

Example Question #2 : Variables

Example Question #1 : Variables

Example Question #2131 : Sat Mathematics

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

Example Question #1 : Direct And Inverse Variation

Example Question #2 : Direct And Inverse Variation

Example Question #8 : Variables

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

All SAT Math Resources

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