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HiSET: Math : HiSet: High School Equivalency Test: Math

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Example Questions

Example Question #4 : Rearrange Formulas/Equations To Highlight A Quantity Of Interest

Solve for :

xy+20 = z

Possible Answers:

$x= \frac{z - 20}{y}$

x=yz - 20

x= z - 20y

$x= \frac{z - 20y}{y}$

$x= \frac{yz - 20}{y}$

Correct answer:

$x= \frac{z - 20}{y}$

Explanation:

To solve for in a literal equation, use the properties of algebra to isolate on one side, just as if you were solving a regular equation.

xy+20 = z

Subtract 20 from both sides:

xy+20- 20 = z - 20

xy = z - 20

Divide both sides by :

$\frac{xy}{y} = \frac{z - 20}{y}$

$x= \frac{z - 20}{y}$ ,

the correct response.

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Example Question #251 : Hi Set: High School Equivalency Test: Math

Solve for :

$32 -x^{2} = y$

You my assume is positive.

Possible Answers:

$x=\sqrt{ y - 32}$

$x=\sqrt{ y}- 32$

$x=\sqrt{ y^{2}- 32}$

$x=\sqrt{ 32- y}$

$x=\sqrt{ 32}- y$

Correct answer:

$x=\sqrt{ 32- y}$

Explanation:

To solve for in a literal equation, use the properties of algebra to isolate on one side, just as if you were solving a regular equation.

First, add $x^{2} - y$ to both sides:

$32 -x^{2} +(x^{2} - y) = y +(x^{2} - y)$

$32 -x^{2} +x^{2} - y = x^{2} +y - y$

$32 - y = x^{2}$

Take the positive square root of both sides:

$x=\sqrt{ 32- y}$ ,

the correct response.

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Example Question #1 : Rearrange Formulas/Equations To Highlight A Quantity Of Interest

Solve for :

$\sqrt{x+1}= y+1$

Possible Answers:

$x = y^{2}+ y$

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

$x = y^{2}- y$

$x = y^{2}+ 2 y$

$x = y^{2}$

Correct answer:

$x = y^{2}+ 2 y$

Explanation:

To solve for in a literal equation, use the properties of algebra to isolate on one side, just as if you were solving a regular equation.

First, square both sides to eliminate the radical symbol:

$\sqrt{x+1}= y+1$

$(\sqrt{x+1})^{2}=( y+1)^{2}$

$x+1=( y+1)^{2}$

Rewrite the expression on the right using the square of a binomial pattern:

$x+1= y^{2}+ 2 (y)(1)+1^{2}$

$x+1= y^{2}+ 2 y+1$

Subtract 1 from both sides:

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

$x = y^{2}+ 2 y$ ,

the correct response.

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Example Question #1 : Functions And Function Notation

Define $f(x)= -\frac{1}{x+1}$ and g(x) = |x+2| .

Evaluate $(f\circ g )(2)$ .

Possible Answers:

The expression is undefined.

$\frac{1}{3}$

$-\frac{1}{4}$

$- \frac{1}{5}$

Correct answer:

$- \frac{1}{5}$

Explanation:

By definition,

$(f\circ g )(2) = f(g(2))$ ,

so first, evaluate g(2) by substituting 2 for in g(x) :

g(x) = |x+2|

g(2) = |2+2| = |4| =4

$(f\circ g )(2) = f(4)$ ,

so evaluate f(4) through a similar substitution:

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

$f(4)= -\frac{1}{4+1} = - \frac{1}{5}$ ,

the correct response.

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Example Question #251 : Hi Set: High School Equivalency Test: Math

Define functions and as follows:

$f(x)= \left\{\begin{matrix} x^{2} -2,&x< 0 \\ x+ 3,& x \ge 0\end{matrix}\right.$

g(x)= x+ 4

Evaluate $(f \circ g)(0)$ .

Possible Answers:

Undefined

Correct answer:

Explanation:

By definition, $(f \circ g)(0) = f(g(0))$ . Evaluate g(0) by substituting 0 for in the definition of :

g(x)= x+ 4

g(0)= 0+ 4= 4

$(f \circ g)(0) = f(g(0)) = f(4)$ , so substitute 4 for in the definition of . Since 4 satisfies the condition $x \ge 0$ , use the definition for those values:

f(x)= x+ 3

f(4)= 4+ 3 = 7 ,

the correct value.

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Example Question #1 : Functions And Function Notation

Consider the scenario below:

Helen is a painter. It takes her 3 days to make each painting. She has already made 6 paintings. Which of the following functions best models the number of paintings she will have after days?

Possible Answers:

f(x)=x-3

f(x)=3x

f(x)=6x^3

$f(x)=\frac{x}{3} +6$

f(x)=3x+6

Correct answer:

$f(x)=\frac{x}{3} +6$

Explanation:

The question asks, "Which of the following functions best models the number of paintings she will have after days?"

From this, you know that the variable represents the number of days, and that f(x) represents the number of paintings she makes as a function of days spent working.

If it takes 3 days to make a painting, each day results in $\frac{1}{3}$ paintings. Therefore, we have a linear relationship with slope $\frac{1}{3}$ .

Additionally, she begins with 6 paintings. Therefore, even when zero days are spent working on paintings, she will have 6 paintings. In other words, f(0)=6 . This means the y-intercept is 6.

As a result, the function will be

$f(x)=\frac{1}{3}x +6$

which can be rewritten as

$f(x)=\frac{x}{3} +6$

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Example Question #2 : Interpret Statements That Use Function Notation In Terms Of A Context

The daily pay in U.S. Dollars for a certain job is defined as the following function f(x) , where equals time in hours:

f(x) = 15x+10

If an employee works for 7 hours in a day, how much is he or she paid?

Possible Answers:

$\$70.00$

$\$25.00$

$\$85.00$

$\$105.00$

$\$115.00$

Correct answer:

$\$115.00$

Explanation:

f(x) = 15x+10

The above function can be understood as, "An employee is paid $15 for each hour he or she works, plus a flat amount of $10."

If an employee works 7 hours, we can find the amount that he or she is paid by plugging in 7 for "Hours" in the equation:

f(x)= 15(7)+10

Adhering to order of operations, we next find the product of 15 and 7:

f(x) = 105+10

Finally, we find the sum of 105 and 10:

f(x) = 115

The employee is paid $115.00 for seven hours' work.

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Example Question #1 : Write A Function That Describes A Relationship Between Two Quantities

f(x) is a linear function defined on the set of real numbers. Four values of are given in the following table:

$\begin{matrix} \underline{x}& \underline{f(x)} \\ 13 & 4 \\ 10 & 6 \\ 7 & 8 \\ 4 & 10 \end{matrix}$

Give the definition for f(x) .

Possible Answers:

$f(x) = -\frac{3}{2} x+19$

f(x) = -2x + 24

$f(x) = -\frac{2}{3} x+14$

$f(x) = \frac{3}{2} x-11$

f(x) = 2x + 5

Correct answer:

$f(x) = -\frac{3}{2} x+19$

Explanation:

By the two-point formula, the equation of a line through points $(x_{1}, y_{1}), (x_{2}, y_{2})$ is

$y-y_{1} = m(x- x_{1})$ ,

where $m = \frac{x_{2}-x_{1}}{y_{2}-y_{1}}$ , the slope of the line. The choice of the two points from the given four is arbitrary, so we will select the middle two ordered pairs. Substituting $x_{1}= 10,y_{1} =6, x_{2}= 7, y_{2}= 8$ , then

$m = \frac{x_{2}-x_{1}}{y_{2}-y_{1}}= \frac{7-10}{8-6} = -\frac{3}{2}$

The first formula becomes

$y-10 = -\frac{3}{2}(x-6)$

Since f(x)= y , solve for . First, distribute $-\frac{3}{2}$ throughout the difference at right:

$y-10 = -\frac{3}{2} x- \left (-\frac{3}{2} \right )6$

$y-10 = -\frac{3}{2} x- \left (-9 \right )$

$y-10 = -\frac{3}{2} x+9$

Isolate by adding 10 to both sides:

$y-10+ 10 = -\frac{3}{2} x+9+ 10$

$y = -\frac{3}{2} x+19$

Replacing, the definition of f(x) is

$f(x) = -\frac{3}{2} x+19$ .

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Example Question #1 : Domain And Range Of A Function

Define functions $f(x) = \sqrt{x}$ and $g(x) =- x^{2}$

Give the domain of the composition $g\circ f$ .

Possible Answers:

$\left ( 0, \infty\right )$

$\left [ 0, \infty\right )$

$\left ( - \infty, 0 \right ]$

$\left ( - \infty, 0 \right )$

None of the other choices gives the correct response.

Correct answer:

$\left [ 0, \infty\right )$

Explanation:

The domain of the composition of functions $g\circ f$ is that the set of all values that fall in the domain of such that f(c) falls in the domain of .

$f(x) = \sqrt{x}$ , a square root function, whose domain is the set of all values of that make the radicand nonnegative. Since the radicand is itself, the domain of is simply the set of all nonnegative numbers, or $\left [ 0, \infty\right )$ .

$g(x) =- x^{2}$ , a polynomial function, whose domain is the set of all real numbers. Therefore, the domain is not restricted further by . The domain of $g\circ f$ is $\left [ 0, \infty\right )$ .

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Example Question #2 : Domain And Range Of A Function

Define functions $f(x) = \frac{1}{x-1}$ and $g(x)= \sqrt{x}+ 1$ .

Give the domain of the composition $g\circ f$ .

Possible Answers:

$(0, 1) \cup (1, \infty)$

$(-\infty, 1)$

(0, 1)

$(1, \infty)$

$(-\infty, 1) \cup (1, \infty)$

Correct answer:

$(1, \infty)$

Explanation:

The domain of the composition of functions $g\circ f$ is that the set of all values that fall in the domain of such that f(c) falls in the domain of .

$f(x) = \frac{1}{x-1}$ , a rational function, so its domain is the set of all numbers except those that make its denominator zero. x-1= 0 if and only if x= 1 , making its domain $(-\infty, 1) \cup (1, \infty)$ .

$g(x)= \sqrt{x}+ 1$ , a square root function, so the domain of is the set of all values that make the radicand a nonnegative number. Since the radicand is itself, the domain of is simply the set of all nonnegative numbers, or $\left [ 0, \infty\right )$ .

Therefore, we need to further restrict the domain to those values in $(-\infty, 1) \cup (1, \infty)$ that make

$f(c) \ge 0$

Equivalently,

$\frac{1}{c-1}\ge 0$

Since 1 is positive, and the quotient is nonnegative as well, it follows that

c-1 > 0 ,

and

c > 1 .

The domain of $g\circ f$ is therefore $(1, \infty)$ .

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HiSET: Math : HiSet: High School Equivalency Test: Math

Study concepts, example questions & explanations for HiSET: Math

All HiSET: Math Resources

Example Questions

Example Question #4 : Rearrange Formulas/Equations To Highlight A Quantity Of Interest

Example Question #251 : Hi Set: High School Equivalency Test: Math

Example Question #1 : Rearrange Formulas/Equations To Highlight A Quantity Of Interest

Example Question #1 : Functions And Function Notation

Example Question #251 : Hi Set: High School Equivalency Test: Math

Example Question #1 : Functions And Function Notation

Example Question #2 : Interpret Statements That Use Function Notation In Terms Of A Context

Example Question #1 : Write A Function That Describes A Relationship Between Two Quantities

Example Question #1 : Domain And Range Of A Function

Example Question #2 : Domain And Range Of A Function

All HiSET: Math Resources

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