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tatiyna
3 years ago
6

What is the inverse of the function f(x) = x – 12?

Mathematics
2 answers:
denis23 [38]3 years ago
6 0
F (x) =x+12 is the inverse of the function
Nostrana [21]3 years ago
3 0

Answer:

f^{-1}(x)=g(x)=(x+12)

Step-by-step explanation:

The given function is f(x) = ( x-12)

To find the inverse we will rewrite the function in an equation form

y = ( x - 12 )

Now we will replace y by x and x by y

x = y - 12

Now we will solve it for the value of y

y = ( x + 12 )

Finally we will rewrite the equation into a function.

g(x) = x + 12

This g (x) is the inverse function of f(x).

f^{-1}(x)=g(x)=(x+12)

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The null hypothesis in this case is

Null hypothesis: H₀: ∪ = ≤ 393 pounds

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4 0
3 years ago
Analyze ady Porter is buying t ride tickets at the county fair. He receives 3 tickets for every dollar, d, that he spends. Which
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Answer:

The answer is below

Step-by-step explanation:

An independent variable is a variable that does not dependent on other variables. It is the input variable.

A dependent variable is a variable that is dependent on other variables. The dependent variable depends on the independent variable. It is the output variable.

Since porter receives 3 tickets for every dollar, this means the number of ticket is dependent on the money he spend. Therefore the money he spends (d) is the independent variable and the tickets (t) is the dependent variable.

4 0
3 years ago
Philip ran 4 3/8 miles yesterday. Michael ran 1 5/8 miles yesterday.
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3 years ago
Yolunda club has 35 members. it rules require that 60% of them be present to vote. at least how many members must be present to
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3 years ago
For number 6, evaluate the definite integral.
maks197457 [2]
\bf \displaystyle \int\limits_{0}^{28}\ \cfrac{1}{\sqrt[3]{(8+2x)^2}}\cdot dx\impliedby \textit{now, let's do some substitution}\\\\
-------------------------------\\\\
u=8+2x\implies \cfrac{du}{dx}=2\implies \cfrac{du}{2}=dx\\\\
-------------------------------\\\\

\bf \displaystyle \int\limits_{0}^{28}\ \cfrac{1}{\sqrt[3]{u^2}}\cdot \cfrac{du}{2}\implies \cfrac{1}{2}\int\limits_{0}^{28}\ u^{-\frac{2}{3}}\cdot du\impliedby 
\begin{array}{llll}
\textit{now let's change the bounds}\\
\textit{by using } u(x)
\end{array}\\\\
-------------------------------\\\\
u(0)=8+2(0)\implies u(0)=8
\\\\\\
u(28)=8+2(28)\implies u(28)=64

\bf \\\\
-------------------------------\\\\
\displaystyle  \cfrac{1}{2}\int\limits_{8}^{64}\ u^{-\frac{2}{3}}\cdot du\implies \cfrac{1}{2}\cdot \cfrac{u^{\frac{1}{3}}}{\frac{1}{3}}\implies \left. \cfrac{3\sqrt[3]{u}}{2} \right]_8^{64}
\\\\\\
\left[ \cfrac{3\sqrt[3]{(2^2)^3}}{2} \right]-\left[ \cfrac{3\sqrt[3]{2^3}}{2}  \right]\implies \cfrac{12}{2}-\cfrac{6}{2}\implies 6-3\implies 3
3 0
3 years ago
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