Question

1. which of the following has an inverse that is a function? (select all that apply)

a)

x	-3	-2	-1	0	1	2	3
y	5	5	5	5	5	5	5

b)

x	-3	-2	-1	0	1	2	3
y	4	3	2	1	0	-1	-2

c) $y = \frac{22}{x}$
d) graph of a v - shaped graph

1. a function is shown below. what is the value of $f(24) - f(0)$?

$f(x)=\

$$\begin{cases}\\frac{1}{2}x - 15, & x\\leq - 4\\\\20 - 3x^{2}, & x > - 4\\end{cases}$$

$

1. the function below $f(x)$ has $f(3)=7$ and $f(4)=49$ as shown. what is the total number of real zeros of $f(x)=2x^{3}-17x^{2}+22x - 192x^{2}+69x + 136$?
2. which expression is equivalent to $\frac{(3x^{2}y^{3})^{2}}{(6x^{4}y)^{3}}$?
3. if $f(x)=5x - 7$, find the inverse of $f(x)$.
4. what value of $n$ makes the equation $3(10)^{n}=27$ true?
5. solve $9 - 5x = 49$. what is the value of $x$?

(other problems partially visible related to functions, polynomials, equations)

Explanation:

Step1: Identify the problem type

This is a set of mathematics problems, likely involving functions, equations, and algebraic manipulations. Let's start with problem 18: finding the value of \( f(24) \) given \( f(x)=\frac{\frac{1}{2}x - 15}{x - 4} \) when \( x = 24 \).

Step2: Substitute \( x = 24 \) into the function

Substitute \( x = 24 \) into \( f(x) \):

$$ f(24)=\frac{\frac{1}{2}(24)-15}{24 - 4} $$

First, calculate the numerator: \( \frac{1}{2}(24)=12 \), then \( 12 - 15=-3 \).
Denominator: \( 24 - 4 = 20 \).
So \( f(24)=\frac{-3}{20}=-0.15 \)? Wait, no, maybe I misread. Wait, the function is \( f(x)=\frac{\frac{1}{2}x - 15}{x - 4} \)? Wait, maybe the original problem is \( f(x)=\frac{\frac{1}{2}x - 15}{x - 4} \) when \( x = 24 \). Wait, let's recalculate:
\( \frac{1}{2}(24)=12 \), \( 12 - 15=-3 \), denominator \( 24 - 4 = 20 \), so \( f(24)=-\frac{3}{20}=-0.15 \). But maybe the function is different. Wait, maybe the numerator is \( \frac{1}{2}x - 15 \) and denominator \( x - 4 \). Alternatively, maybe a typo, but let's proceed.

For problem 19: The function has \( f(x)=26 - 13x \) (wait, the original says "The function below has \( f(x)=26 - 13x \) (and \( x \) and 49 as shown). What is the total number of real zeros of \( f(x)=26 - 13x \), \( g(x)=192x^2 - 169x + 39 \)". Wait, linear function \( f(x)=26 - 13x \) is a line, so it has one real zero (when \( 26 - 13x = 0 \Rightarrow x = 2 \)). For \( g(x)=192x^2 - 169x + 39 \), we can use discriminant \( D = b^2 - 4ac \), where \( a = 192 \), \( b=-169 \), \( c = 39 \). \( D=(-169)^2 - 4\times192\times39 \). Calculate \( 169^2 = 28561 \), \( 4\times192\times39 = 4\times7488 = 29952 \). So \( D = 28561 - 29952 = -1391 \), which is negative, so \( g(x) \) has no real zeros. So total real zeros: 1 (from \( f(x) \)) + 0 (from \( g(x) \)) = 1? Wait, maybe I misread the functions.

Problem 20: Simplify expressions. Let's take option a: \( \frac{1}{x^{-2}} = x^2 \) (since \( \frac{1}{a^{-n}} = a^n \)). Option b: \( \frac{9x^3}{3x} = 3x^2 \) (divide coefficients: 9/3=3, subtract exponents: \( x^3/x = x^{3 - 1}=x^2 \)). Option c: \( \frac{(x^3)^2}{x^4}=\frac{x^6}{x^4}=x^{6 - 4}=x^2 \). Option d: \( \frac{(x + 1)^2}{x + 1}=x + 1 \) (as long as \( x
eq - 1 \)). So need to check which is equivalent.

Problem 21: Find the inverse of \( f(x)=5x - 7 \). Let \( y = 5x - 7 \), solve for \( x \): \( y + 7 = 5x \Rightarrow x=\frac{y + 7}{5}=\frac{1}{5}y+\frac{7}{5} \), so \( f^{-1}(x)=\frac{1}{5}x+\frac{7}{5} \), which is option b?

Problem 22: Solve \( 3(0.10)^x = 27 \). Divide both sides by 3: \( (0.10)^x = 9 \). Take log: \( x\log(0.10)=\log(9) \Rightarrow x=\frac{\log(9)}{\log(0.10)}=\frac{\log(9)}{-1}\approx - 0.954 \), but options are 0.10, 0.13, 0.19, 0.44. Wait, maybe the equation is \( 3(10)^x = 27 \)? Then \( 10^x = 9 \Rightarrow x=\log(9)\approx0.954 \), no. Wait, maybe \( 3(0.10)^x = 27 \) is wrong, maybe \( 3(10)^{-x}=27 \), so \( 10^{-x}=9 \Rightarrow -x=\log(9)\Rightarrow x = -\log(9)\approx - 0.954 \), not matching. Maybe a typo.

Problem 23: Solve \( 9 - 5x = 49 \). Subtract 9: \( - 5x = 40 \Rightarrow x=-8 \). Then find the value of \( 6x \): \( 6\times(-8)=-48 \). So the value is -48, which is option a?

(Problem 18):

Step1: Substitute \( x = 24 \) into \( f(x) \)

Given \( f(x)=\frac{\frac{1}{2}x - 15}{x - 4} \), substitute \( x = 24 \):

$$ f(24)=\frac{\frac{1}{2}(24)-15}{24 - 4} $$

Step2: Calculate numerator and denominator

Numerator: \( \frac{1}{2}(24)=12 \), \( 12 - 15=-3 \)
Denominator: \( 24 - 4 = 20 \)

Step3: Simplify the fraction

$$ f(24)=\frac{-3}{20}=-0.15 $$

Answer:

For problem 18: \( f(24)=\frac{\frac{1}{2}(24)-15}{24 - 4}=\frac{12 - 15}{20}=\frac{-3}{20}=-0.15 \)

For problem 19: \( f(x)=26 - 13x \) has one zero (\( x = 2 \)), \( g(x)=192x^2 - 169x + 39 \) has discriminant \( D = (-169)^2 - 4\times192\times39 = 28561 - 29952=-1391<0 \), so 0 zeros. Total: 1.

For problem 20: Depends on the options, but \( \frac{(x + 1)^2}{x + 1}=x + 1 \) (d), \( \frac{9x^3}{3x}=3x^2 \) (b), etc.

For problem 21: \( f^{-1}(x)=\frac{1}{5}x+\frac{7}{5} \) (option b)

For problem 22: Maybe a misprint, but if solving \( 3(10)^x = 27 \), \( x=\log(9)\approx0.954 \), not matching. If \( 3(0.1)^x = 27 \), \( x=\log(9)/\log(0.1)\approx - 0.954 \), no.

For problem 23: \( x=-8 \), \( 6x=-48 \), so answer is -48 (option a: 0.0794? No, maybe option a is -48? Wait, options are a) 0.0794, b) 0.0995, c) 0.2443, d) 0.474. No, maybe miscalculation. Wait, \( 9 - 5x = 49 \Rightarrow - 5x = 40 \Rightarrow x=-8 \), then \( 6^x = 6^{-8}=\frac{1}{6^8}\approx\frac{1}{1679616}\approx0.000000595 \), no. Wait, maybe the question is "What is the value of \( 6^x \) when \( x=-8 \)", so \( 6^{-8}=\frac{1}{6^8}\approx0.000000595 \), not matching. Maybe the equation is \( 9 + 5x = 49 \), then \( 5x = 40 \Rightarrow x = 8 \), \( 6^8 = 1679616 \), no. Maybe the original problem is different.

Given the complexity, let's focus on problem 18: