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3 years ago

6

Find the domain of:

e="f(x) = \sqrt{12 - 3x} " alt="f(x) = \sqrt{12 - 3x} " align="absmiddle" class="latex-formula">
Plz help

2 answers:

ch4aika [34]3 years ago

7 0

The domain is 4 that’s your answer

Alexxandr [17]3 years ago

3 0

Answer:

Step-by-step explanation:

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The students at a community center are participating in a bike riding fundraiser at a local park. A local company is donating $1

Oksi-84 [34.3K]

Answer:

This can not be properly answered, because we do not have the length of each trail.

The first step here will be find the length, in miles, for each of the four trails.

As students choose all four trails, the number of miles that each student will bike is equal to the addition of the lenght of the four trails.

Suppose this number is N.

So each student bike N miles, and the company donates $1.75 per mile. Then the amount of money that a single student is N times $1.75, or N*$1.75

5 0

3 years ago

4TH TIME ASKING THIS!!! Please help me! Someone pleaseeee. I need the correct answers. I don’t want to fail

Alexeev081 [22]

Answer:

The functions are inverses; f(g(x)) = x ⇒ answer D

$h^{-1}(x)=\sqrt{\frac{x+1}{3}}$ ⇒ answer D

Step-by-step explanation:

* <em>Lets explain how to find the inverse of a function</em>

- Let f(x) = y

- Exchange x and y

- Solve to find the new y

- The new y = $f^{-1}(x)$

* <em>Lets use these steps to solve the problems</em>

∵ $f(x)=\sqrt{x-3}$

∵ f(x) = y

∴ $y=\sqrt{x-3}$

- Exchange x and y

∴ $x=\sqrt{y-3}$

- Square the two sides

∴ x² = y - 3

- Add 3 to both sides

∴ x² + 3 = y

- Change y by $f^{-1}(x)$

∴ $f^{-1}(x)=x^{2}+3$

∵ g(x) = x² + 3

∴ $f^{-1}(x)=g(x)$

∴ <u><em>The functions are inverses to each other</em></u>

* <em>Now lets find f(g(x))</em>

- To find f(g(x)) substitute x in f(x) by g(x)

∵ $f(x)=\sqrt{x-3}$

∵ g(x) = x² + 3

∴ $f(g(x))=\sqrt{(x^{2}+3)-3}=\sqrt{x^{2}+3-3}=\sqrt{x^{2}}=x$

∴ <u><em>f(g(x)) = x</em></u>

∴ The functions are inverses; f(g(x)) = x

* <em>Lets find the inverse of h(x)</em>

∵ h(x) = 3x² - 1 where x ≥ 0

- Let h(x) = y

∴ y = 3x² - 1

- Exchange x and y

∴ x = 3y² - 1

- Add 1 to both sides

∴ x + 1 = 3y²

- Divide both sides by 3

∴ $\frac{x + 1}{3}=y^{2}$

- Take √ for both sides

∴ ± $\sqrt{\frac{x+1}{3}}=y$

∵ x ≥ 0

∴ We will chose the positive value of the square root

∴ $\sqrt{\frac{x+1}{3}}=y$

- replace y by $h^{-1}(x)$

∴ $h^{-1}(x)=\sqrt{\frac{x+1}{3}}$

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