<img src="https://tex.z-dn.net/?f=g%28x%29%20%3D%203x%20-%206" id="TexFormula1" title="g(x) = 3x - 6" alt="g(x) = 3x - 6" align=

Ask question

Ask question

All categories

3 years ago

10

"absmiddle" class="latex-formula">and
${g}^{ - 1}(x) = \frac{x + 6}{3}$ . With g(x) and
${g}^{ - 1}(x)$ , find
${g}^{ - 1}(g(10))$ . Make sure to show your work.

1 answer:

VARVARA [1.3K]3 years ago

3 0

Hey there! :)

Answer:

$g^{-1} (g(10))=10$

Step-by-step explanation:

Begin by calculating g(10):

g(x) = 3x - 6

Substitute in 10 for x:

g(10) = 3(10) - 6

g(10) = 30-6

g(10) = 24.

Plug '24' into 'x' into $g^{-1} (x)$

$g^{-1} (24)=\frac{(24) + 6}{3}$

Simplify:

$g^{-1} (24)=\frac{30}{3}$

$g^{-1} (24)=10$

You might be interested in

Which graphs show direct variation chech all that apply

kirill [66]

Answer:

Is there a picture you are missing? We cant answer without knowing what the answer choices are.

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

0.3 km = ? mi. rounded to the nearest hundred

Sergio [31]

Answer:

0.19

Step-by-step explanation:

There are approximately .62 miles in a kilometer

6 0

3 years ago

Read 2 more answers

Find the median of the following set of data. Round your answer to the nearest hundredth, if necessary.

Usimov [2.4K]

The correct answer is 35.2 because when you arrange them in order from least to greatest, the middle is 35.2

8 0

3 years ago

What curve passes through the point (1 ,2) and has an arc length on the interval [2,6] given by Integral from 2 to 6 StartRo

Georgia [21]

Answer:

Step-by-step explanation:

Given

Length of curve

$L=\int_{2}^{6}\sqrt{1+64x^{-6}}dx$

Length of curve is given by

$L=\int_{a}^{b}\sqrt{1+\left ( \frac{\mathrm{d} y}{\mathrm{d} x}\right )^2}dx$ over interval a to b

comparing two we get

$\frac{\mathrm{d} y}{\mathrm{d} x}=8x^{-3}$

$dy=8x^{-3}dx$

integrating

$\int dy=\int 8x^{-3}dx$

$y=-4x^{-2}+C$

Curve Passes through (1,2)

$1=-4+C$

$C=5$

curve is

$y+\frac{4}{x^2}=5$

3 0

3 years ago

X^2 - y^2 = 100

luda_lava [24]

Answer:

20

Step-by-step explanation:

Let's start by rewriting the second equation in terms of "x":

$x+y=5$

Subtract y from both sides:

$x=5-y$

Now, substitute "5-y" for "x" in the first equation:

$(5-y)^2-y^2=100$

Note that:

$(a-b)^2=a^2-2ab+b^2$

$25-10y+y^2-y^2=100$

Cancel out like terms:

$25-10y=100$

Subtract 25 from both sides:

$-10y=75$

Divide both sides by -10

$y=\frac{75}{-10}=\frac{15}{-2}=-\frac{15}{2}$

Now, substitute this value back into either of the equations to solve for x.

$x+y=5\\x-\frac{15}{2}=5\\$

Add 15/2 to both sides:

$x=5+\frac{15}{2}\\x=\frac{10}{2}+\frac{15}{2}\\x=\frac{25}{2}$

Now, find the difference:

$x-y=\frac{25}{2}-(-\frac{15}{2})=\frac{25}{2}+\frac{15}{2}=\frac{40}{2}=20$

8 0

2 years ago