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SAT II Math I : SAT Subject Test in Math I

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Example Questions

Example Question #17 : Simplifying Expressions

Exponentiate:

$\left ( 9R - 10\right )^{3}$

Possible Answers:

$729R^{3}-2,430R^{2} - 2,700R+1,000$

$729R^{3}-2,430R^{2} + 2,700R-1,000$

$729R^{3}-810R^{2} + 900R-1,000$

$729R^{3}-1,000$

$729R^{3}-810R^{2} - 900R+1,000$

Correct answer:

$729R^{3}-2,430R^{2} + 2,700R-1,000$

Explanation:

The difference of two terms can be cubed using the pattern

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

Where A = 9R, B = 10 :

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

$(9R-10)^{3} = (9R) ^{3}-3(9R)^{2}\cdot 10 + 3(9R) \cdot 10^{2}-10^{3}$

$(9R-10)^{3} =729R^{3}-3\cdot 81R^{2}\cdot 10 + 3(9R) \cdot 100-1,000$

$(9R-10)^{3} =729R^{3}-2,430R^{2} + 2,700R-1,000$

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Example Question #18 : Simplifying Expressions

How many of the following are prime factors of $81y^{4}- 1$ ?

I) $9y^{2} + 1$

II) $3y^{2} + 1$

III) 3y+1

IV) 3y-1

Possible Answers:

None

One

Four

Three

Two

Correct answer:

Three

Explanation:

Factor $y^{4}- 81$ all the way to its prime factorization.

$y^{4}- 81$ can be factored as the difference of two perfect square terms as follows:

$y^{4}- 81$

$=\left ( 9y^{2} \right )^{2} - 1^{2}$

$=\left (9 y^{2} + 1 \right )\left ( 9y^{2} - 1 \right )$

$9y^{2} + 1$ is a factor, and, as the sum of squares, it is a prime. $9y^{2} - 1$ is also a factor, but it is not a prime factor - it can be factored as the difference of two perfect square terms. We continue:

$\left (9 y^{2} + 1 \right )\left ( 9y^{2} - 1 \right )$

$=\left ( 9y^{2} + 1 \right )\left [ \left (3y \right ) ^{2} - 1^{2} \right ]$

$=\left ( 9y^{2} + 1 \right )\left (3y+1 \right )\left (3 y-1\right )$

Therefore, of the given four choices, only $3y^{2} + 1$ is not a factor, so the correct response is three.

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Example Question #4 : How To Find The Solution Of A Rational Equation With A Binomial Denominator

$\small \frac{x^2-5x-6}{x+1}$

For all values $x\neq -1$ , which of the following is equivalent to the expression above?

Possible Answers:

$\small x+6$

$\small x-3$

$\small x-6$

$\small x-2$

Correct answer:

$\small x-6$

Explanation:

First, factor the numerator. We need factors that multiply to $\small -6$ and add to $\small -5$ .

$\small 1*-6=-6\ \text{and}\ 1+(-6)=-5$

$\small x^2-5x-6=(x-6)(x+1)$

We can plug the factored terms into the original expression.

$\small \frac{x^2-5x-6}{x+1}=\frac{(x-6)(x+1)}{(x+1)}$

Note that $\small (x+1)$ appears in both the numerator and the denominator. This allows us to cancel the terms.

$\frac{(x-6)(x+1)}{(x+1)}=(x-6)$

This is our final answer.

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Example Question #13 : Simplifying Expressions

Simplify the following expression:

3x+18y=81

Possible Answers:

2x+9y=40

x+18y=81

3x+6y=27

x+6y=27

x+6y=81

Correct answer:

x+6y=27

Explanation:

When simplifying an equation,you must find a common factor for all values in the equation, including both sides.

3,18, and, can all be divided by so divide them all at once

$\left ( \frac{3}{3}=1, \frac{18}{3}=6, \frac{81}{3}=27\right )$ .

This leaves you with

x+6y=27 .

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Example Question #21 : Simplifying Expressions

Simplify the expression $\frac{x^5(3x^2y^3)}{x^4y^2z}$

Possible Answers:

Already in simplest form

$\frac{3x^6y}{z}$

$\frac{3x^3y}{z}$

$\frac{3x^3}{yz}$

3x^3y

Correct answer:

$\frac{3x^3y}{z}$

Explanation:

$\frac{x^5(3x^2y^3)}{x^4y^2z}$

Simplify the numerator by multiplying in the $\tiny x^5$ term

$\frac{3x^7y^3}{x^4y^2z}$

Cancel out like terms in the numerator and denominator.

$\frac{3x^3y}{z}$

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Example Question #22 : Simplifying Expressions

Simplify: $(x^2)\cdot (\frac{x^{-1}}{x})$

Possible Answers:

x^3

$\frac{1}{x^3}$

$\frac{1}{x}$

$\frac{1}{x^2}$

Correct answer:

Explanation:

Rewrite the denominator of the second fraction using a power.

$(\frac{x^{-1}}{x}) = (\frac{x^{-1}}{x^1})$

Using the subtraction rule of exponents, we can simplify this as one term.

$x^{-1-1} = x^{-2}$

The expression becomes:

$(x^2)\cdot (x^{-2})$

Apply the addition rule of exponents.

$x^{2+(-2)} = x^0 =1$

The answer is:

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Example Question #23 : Simplifying Expressions

$\frac{(x^7 - x^6)}{ (x - 1)} =$

Possible Answers:

$\frac{1}{x-1}$

x^6

x^5

x^6 - x^5

$\frac{x}{(x-1)}$

Correct answer:

x^6

Explanation:

Factor an x^6 in the numerator to get:

$\frac{(x^7 - x^6)}{ (x - 1)} =\frac{x^6(x-1)}{(x-1)}$

We can now cancel out (x-1) from the numerator and denominator leaving the answer.

x^6

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Example Question #1 : Range And Domain

Define $f (x) = \sqrt{25-x^{2}}$ .

Give the range of .

Possible Answers:

[0, 25]

$(-\infty, 5]$

[-25,25]

[0, 5]

[-5,5]

Correct answer:

[0, 5]

Explanation:

The radicand within a square root symbol must be nonnegative, so

$25 - x^{2} \geq 0$

$25 \geq x^{2}$

$x^{2} \leq 25$

This happens if and only if $-5 \leq x \leq 5$ , so the domain of is [-5,5] .

$f (x) = \sqrt{25-x^{2}}$ assumes its greatest value when $25-x^{2}$ , which is the point on [-5,5] where $x^{2}$ is least - this is at x = 0 .

$f (0) = \sqrt{25-0^{2}} = \sqrt{25} = 5$

Similarly, $f (x) = \sqrt{25-x^{2}}$ assumes its least value when $25-x^{2}$ , which is the point on [-5,5] where $x^{2}$ is greatest - this is at $x = \pm 5$ .

$f (5) = \sqrt{25-5^{2}} = \sqrt{0} = 0$

$f (-5) = \sqrt{25-\left (-5 \right )^{2}} = \sqrt{0} = 0$

Therefore, the range of is [0, 5] .

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Example Question #1 : Range And Domain

Define $f (x) = \sqrt{16-x^{2}}$ .

Give the domain of .

Possible Answers:

[0, 4]

$[-4, \infty)$

$(-\infty, 4]$

$(-\infty, \infty)$

[-4,4]

Correct answer:

[-4,4]

Explanation:

The radicand within a square root symbol must be nonnegative, so

$16 - x^{2} \geq 0$

$16 \geq x^{2}$

$x^{2} \leq 16$

This happens if and only if $-4 \leq x \leq 4$ , so the domain of is [-4,4] .

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Example Question #2 : Range And Domain

Define the functions and on the set of real numbers as follows:

$f (x) = \sqrt{x- 4}$

$g(x) = x^{2} + 8$

Give the natural domain of the composite function $f \circ g$ .

Possible Answers:

$\left [ 2, \infty \right )$

$\left [-2, \infty \right )$

$\left [ 2\sqrt{2} , \infty \right )$

The set of all real numbers

$\left [-2\sqrt{2} , \infty \right )$

Correct answer:

The set of all real numbers

Explanation:

The natural domain of the composite function $f \circ g$ is defined to be the intersection two sets.

One set is the natural domain of . Since is a polynomial, its domain is the set of all real numbers.

The other set is the set of all values of such that that g(x) is in the domain of . Since the radicand of the square root in must be nonnegative,

$x- 4 \geq 0$ , and $x \geq 4$ , the domain of

Therefore, the other set is the set of all such that

$g(x) \geq 4$

Substitute:

$x^{2}+8 \geq 4$

$x^{2} \geq -4$

This holds for all real numbers, so this set is also the set of all real numbers.

The natural domain of $f \circ g$ is the set of all real numbers.

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SAT II Math I : SAT Subject Test in Math I

Study concepts, example questions & explanations for SAT II Math I

All SAT II Math I Resources

Example Questions

Example Question #17 : Simplifying Expressions

Example Question #18 : Simplifying Expressions

Example Question #4 : How To Find The Solution Of A Rational Equation With A Binomial Denominator

Example Question #13 : Simplifying Expressions

Example Question #21 : Simplifying Expressions

Example Question #22 : Simplifying Expressions

Example Question #23 : Simplifying Expressions

Example Question #1 : Range And Domain

Example Question #1 : Range And Domain

Example Question #2 : Range And Domain

All SAT II Math I Resources

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