Single-Variable Algebra - SAT Subject Test in Math II

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Question

Adult tickets to the zoo sell for ; child tickets sell for . On a given day, the zoo sold $5\textup,450$ tickets and raised $$36\textup,330$ in admissions. How many adult tickets were sold?

Answer

Let be the number of adult tickets sold. Then the number of child tickets sold is $5\textup,450 - A$ .

The amount of money raised from adult tickets is ; the amount of money raised from child tickets is $5 \left (5\textup,450 - A \right )$ . The sum of these money amounts is , so the amount of money raised can be defined by the following equation:

$9A + 5 (5\textup,450-A) = 36\textup,330$

To find the number of adult tickets sold, solve for :

$9A + 27\textup,250-5A = 36\textup,330$

$4A + 27\textup,250 = 36\textup,330$

$4A + 27\textup,250 - 27\textup,250= 36\textup,330- 27\textup,250$

$4A = 9,080\textup$

$4A \div 4 = 9\textup,080 \div 4$

$A = 2\textup,270$

$2\textup,270$ adult tickets were sold.

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Question

Which of the following phrases can be written as the algebraic expression $\left |8-a \right |$ ?

Answer

$\left |8-a \right |$ is the absolute value of , which is the difference of eight and a number. Therefore, $\left |8-a \right |$ is "the absolute value of the difference of eight and a number."

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Question

Which of the following phrases can be written as the algebraic expression ?

Answer

is seven decreased by , which is the opposite of a number; therefore, is "seven decreased by the opposite of a number."

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Question

Which of the following phrases can be represented by the algebraic expression $\frac{1}{9 - x}$ ?

Answer

$\frac{1}{9 - x}$ is the multiplicative inverse of , which is the difference of nine and a number. Therefore, $\frac{1}{9 - x}$ is "the multiplicative inverse of the difference of nine and a number".

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Question

Which of the following phrases can be represented by the algebraic expression

$\sqrt[3]{x} - 10$

Answer

$\sqrt[3]{x} - 10$ is ten less than $\sqrt[3]{x}$ , which is the cube root of a number; therefore, $\sqrt[3]{x} - 10$ is "ten less than the cube root of a number".

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Question

Which of the following phrases can be represented by the algebraic expression

$20 - \sqrt{x}$

Answer

$20 - \sqrt{x}$ is twenty decreased by $\sqrt{x}$ , which is the square root of a number, so $20 - \sqrt{x}$ is "twenty decreased by the square root of a number".

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Question

Sarah sells lemonade at the concession stands. She charges fifty cents per cup of lemonade, and twenty five cents for refills. What is the equation that represents the total that she will make from the lemonade stand using the variables cups and refills ?

Answer

Sarah charges fifty cents per cup of lemonade:

Sarah charges twenty five cents for refills:

Set up the equation by adding the totals.

The answer is:

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Question

Multiply the expressions:

$\small (x^{2} - 2x + 7) (x^{2} - 2x - 7)$

Answer

You can look at this as the sum of two expressions multiplied by the difference of the same two expressions. Use the pattern

$\small (A+B)(A-B) = (A^{2}-B^{2})$ ,

where $\small A = x^{2} - 2x$ and $\small B = 7$ .

$\small \small \small (x^{2} - 2x + 7) (x^{2} - 2x - 7) = (x^{2} - 2x )^{2} - 7^{2} = (x^{2} - 2x )^{2} - 49$

To find $\small (x^{2} - 2x )^{2}$ , you use the formula for perfect squares:

$\small (A-B)^{2} = (A^{2}-2AB+B^{2})$ ,

where $\small \small A = x^{2}$ and $\small B = 2x$ .

$\small \small \small (x^{2}-2x)^{2} = ((x^{2})^{2}-2x^{2}(2x)+(2x)^{2}) = x^{4}-4x^{3}+4x^2$

Substituting above, the final answer is $\small x^{4}-4x^{3}+4x^2-49$ .

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Question

Expand the expression by multiplying the terms.

$\small (x-4)(x+2)(2x-5)$

Answer

$\small (x-4)(x+2)(2x-5)$

When multiplying, the order in which you multiply does not matter. Let's start with the first two monomials.

Use FOIL to expand.

Now we need to multiply the third monomial.

$\small (x-4)(x+2)(2x-5)=(x^2-2x-8)(2x-5)$

Similar to FOIL, we need to multiply each combination of terms.

Combine like terms.

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Question

Which of the following values of would make

$3x^{2} +Nx + 32$

a prime polynomial?

Answer

A polynomial of the form $ax^{2} +bx + c$ whose terms do not have a common factor, such as this, can be factored by rewriting it as $ax^{2} +fx + gx + c$ such that and ; the grouping method can be used on this new polynomial.

Therefore, for $3x^{2} +Nx + 32$ to be factorable, must be the sum of the two integers of a factor pair of $3 \times 32 = 96$ . We are looking for a value of that is not a sum of two such factors.

The factor pairs of 96, along with their sums, are:

1 and 96 - sum 97

2 and 48 - sum 50

3 and 32 - sum 35

4 and 24 - sum 28

6 and 16 - sum 22

8 and 12 - sum 20

Of the given choices, only 30 does not appear among these sums; it is the correct choice.

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Question

How many of the following are prime factors of the polynomial $y^{3} - 125$ ?

(A)

(B)

(C) $y^{2}+ 5y + 2 5$

(D) $y^{2}-5y + 2 5$

Answer

$y^{3} - 125 = y^{3} - 5 ^{3}$

making this polynomial the difference of two cubes.

As such, $y^{3} - 125$ can be factored using the pattern

$A ^{3} - B ^{3} = (A - B) (A^{2}+ AB + B^{2})$

so

$y^{3} - 125 = y^{3} - 5 ^{3} = (y - 5) (y^{2}+ y \cdot 5 + 5^{2})= (y - 5) (y^{2}+ 5y + 2 5 )$

(A) and (C) are both factors, but not (B) or (D), so the correct response is two.

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Question

Factor the trinomial.

$6x^{2}-31x+35$

Answer

Use the -method to split the middle term into the sum of two terms whose coefficients have sum and product . These two numbers can be found, using trial and error, to be and .

and

Now we know that $6x^{2}-31x+35$ is equal to $6x^{2}-10x-21x+35$ .

Factor by grouping.

$(6x^{2}-10x)-(21x-35)$

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Question

Factor completely:

$t^{3} + 343$

Answer

Since both terms are perfect cubes $\left (343 = 7^{3} \right )$ , the factoring pattern we are looking to take advantage of is the sum of cubes pattern. This pattern is

$A^{3}+ B^{3} = (A + B) (A^{2}-AB +B^{2})$

We substitute for and 7 for :

$t^{3} + 343 = t^{3} + 7^{3}$

$= (t+ 7) (t^{2}-t \cdot 7+7^{2})$

$= (t+ 7) (t^{2}-7t+49)$

The latter factor cannot be factored further, since we would need to find two integers whose product is 49 and whose sum is ; they do not exist. This is as far as we can go with the factoring.

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Question

Which of the following values of would make

$t^{6}+N$

a prime polynomial?

Answer

$t^{6}$ is the cube of $t^{2}$ . Therefore, if is a perfect cube, the expression $t^{6}+N$ is factorable as the sum of two cubes. All four of the choices are perfect cubes - 8, 27, 64, and 125 are the cubes of 2, 3, 4 and 5, respectively. The correct response is that none of the choices are correct.

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Question

Which of the following values of would not make

$9t^{2}+N$

a prime polynomial?

Answer

$9t^{2}$ is a perfect square term - it is equal to $\left (3t \right )^{2}$ . All of the values of given in the choices are perfect squares - 25, 36, 49, and 64 are the squares of 5, 6, 7, and 8, respectively.

Therefore, for each given value of , the polynomial is the sum of squares, which is normally a prime polynomial. However, if - and only in this case - the polynomial can be factored as follows:

$9t^{2}+36 = 9 \cdot t^{2} + 9 \cdot 4 = 9 (t^{2}+ 4)$ .

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Question

Give the set of all real solutions of the equation $2x^{4} - 13x^{2} +21 = 0$ .

Answer

Set $y = x^{2}$ . Then $x^{4 } = \left (x^{2} \right )^{2} = y ^{2}$ .

$2x^{4} - 13x^{2} +21 = 0$ can be rewritten as

$2 \left (x^{2} \right )^{2} - 13x^{2} +21 = 0$

Substituting $y^{2}$ for $x^{4 }$ and for $x^{2}$ , the equation becomes

$2 y^{2} - 13y +21 = 0$ ,

a quadratic equation in .

This can be solved using the method. We are looking for two integers whose sum is and whose product is . Through some trial and error, the integers are found to be and , so the above equation can be rewritten, and solved using grouping, as

$2 y^{2} - 7y - 6y +21 = 0$

$(2 y^{2} - 7y )- (6y -21 )= 0$

By the Zero Product Principle, one of these factors is equal to zero:

Either:

Substituting $x^{2}$ back for :

$x^{2} = 3$

Taking the positive and negative square roots of both sides:

$x = \pm \sqrt{3}$ .

Or:

$\frac{2y}{2} =\frac{ 7}{2}$

$y =\frac{ 7}{2}$

Substituting back:

$x^{2} =\frac{ 7}{2}$

Taking the positive and negative square roots of both sides, and applying the Quotient of Radicals property, then simplifying by rationalizing the denominator:

$x =\pm \sqrt{\frac{ 7}{2}} = \pm \frac{\sqrt{7}}{\sqrt{2}} = \pm \frac{\sqrt{7} \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2} } =\pm \frac{\sqrt{14}}{2}$

The solution set is $\left { - \frac{\sqrt{14}}{2}, - \sqrt{3}, \sqrt{3} ,\frac{ \sqrt{14}}{2} \right }$ .

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Question

Which of the following is a factor of the polynomial $x^{4} - 11x^{2} + 23x - 18$ ?

Answer

Call $P(x) = x^{4} - 11x^{2} + 23x - 18$ .

By the Rational Zeroes Theorem, since has only integer coefficients, any rational solution of must be a factor of 18 divided by a factor of 1 - positive or negative. 18 has as its factors 1, 2, 3, 6, 9, and 18; 1 has only itself as a factor. Therefore, the rational solutions of must be chosen from this set:

$\left { 1, 2, 3, 6, 9, 18 -1,- 2,- 3,- 6, -9, -18 \right }$ .

By the Factor Theorem, a polynomial is divisible by if and only if - that is, if is a zero. By the preceding result, we can immediately eliminate and as factors, since 4 and 5 have been eliminated as possible zeroes.

Of the three remaining choices, we can demonstrate that is the factor by evaluating :

$P(x) = x^{4} - 11x^{2} + 23x - 18$

$P(2) = 2^{4} - 11 \cdot 2 ^{2} + 23 \cdot 2 - 18$

$= 16 - 11 \cdot 4 + 23 \cdot 2 - 18$

By the Factor Theorem, it follows that is a factor.

As for the other two, we can confirm that neither is a factor by evaluating and :

$P(1) = 1^{4} - 11 \cdot 1 ^{2} + 23 \cdot 1 - 18$

$= 1- 11 \cdot 1 + 23 \cdot 1 - 18$

$P(3) = 3^{4} - 11 \cdot 3 ^{2} + 23 \cdot 3 - 18$

$= 81 -11\cdot 9 + 23 \cdot 3 - 18$

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Question

Define a function $f(x) = x^{2} + \cos 3x$ .

for exactly one real value of on the interval .

Which of the following statements is correct about ?

Answer

Define $g(x)= f(x) - 4 = x^{2} + \cos 3x - 4$ . Then, if , it follows that .

By the Intermediate Value Theorem (IVT), if is a continuous function, and and are of unlike sign, then for some $c \in (a, b)$ . is a continuous function, so the IVT applies here.

Evaluate for each of the following values: $\left { 1.5, 1.6, 1.7,1.8, 1.9, 2.0 \right }$

$g(x)= x^{2} + \cos 3x - 4$

$g(1.5)= 1.5^{2} + \cos 3 (1.5) - 4$

$= 2.25 + \cos 4.5 - 4$

$\approx 2.25 + (-0.21 )- 4$

$\approx - 1.96$

$g(1.6)= 1.6^{2} + \cos 3 (1.6) - 4$

$= 2.56 + \cos 4.8 - 4$

$\approx 2.56 +0.09 - 4$

$\approx - 1.35$

$g(1.7)= 1.7^{2} + \cos 3 (1.7) - 4$

$=2.89 + \cos 5.1- 4$

$\approx 2.89 + 0.38- 4$

$\approx - 0.73$

$g(1.8)= 1.8^{2} + \cos 3 (1.8) - 4$

$= 3.24 + \cos 5.4 - 4$

$\approx 3.24 + 0.63 - 4$

$\approx - 0.13$

$g(1.9)= 1.9^{2} + \cos 3 (1.9) - 4$

$=3.61 + \cos 5.7 - 4$

$\approx 3.61 + 0.83 - 4$

$\approx 0.44$

$g(2.0)= 2.0^{2} + \cos 3 (2.0) - 4$

$= 4 + \cos 6 - 4$

$\approx 4 +0.96 - 4$

$\approx 0.96$

Only in the case of does it hold that assumes a different sign at each endpoint - . By the IVT, , and , for some $c \in (1.8, 1.9)$ .

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Question

A cubic polynomial with rational coefficients whose lead term is $x^{3}$ has and as two of its zeroes. Which of the following is this polynomial?

Answer

A polynomial with rational coefficients has its imaginary zeroes in conjugate pairs. Two imaginary zeroes are given that are each other's complex conjugate - and . Since the polynomial is cubic - of degree 3 - it has one other zero, which must be real. However, no information is given about that zero. Therefore, the polynomial cannot be determined.

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Question

Define functions and $q (x) = 5^{x}$ .

for exactly one value of on the interval . Which of the following is true of ?

Answer

Define

$= 7x- 5^{x}$

Then if ,

it follows that

,

or, equivalently,

.

By the Intermediate Value Theorem (IVT), if is a continuous function, and and are of unlike sign, then for some $c \in (a, b)$ .

Since polynomial and exponential function are continuous everywhere, so is , so the IVT applies here.

Evaluate for each of the following values: $\left { 1. 0 , 1.1, 1.2, 1.3, 1.4, 1.5 \right }$ :

$g(x) = 7x- 5^{x}$

$g(1) = 7 (1)- 5^{1} = 7 - 5 = 2$

$g(1.1) = 7 (1.1)- 5^{1.1} \approx 7.7 - 5.9 \approx 1.8$

$g(1.2) = 7 (1.2)- 5^{1.2} \approx 8.4 - 6.9 \approx 1.5$

$g(1.3) = 7 (1.3)- 5^{1.3} \approx 9.1 - 8.1 \approx 1$

$g(1.4) = 7 (1.4)- 5^{1.4} \approx 9.8 - 9.5 \approx 0.3$

$g(1.5) = 7 (1.5)- 5^{1.5} \approx 10.5- 11.2 \approx - 0.7$

Only in the case of does it hold that assumes a different sign at both endpoints - . By the IVT, , and , for some $c \in (1.4, 1.5)$ .

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