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3 years ago

Given: f(x) = x - 7 and h(x) = 2x + 3

Mathematics

2 answers:

Vinil7 [7]3 years ago

4 0

Answer:

h(f(x)) = 2x - 11

(the second answer is f(h(x)) = 2x - 4)

Step-by-step explanation:

h(x-7) = 2(x-7) + 3

h(x) = 2x -14 + 3

h(x) = 2x - 11

Andrei [34K]3 years ago

3 0

<u>Answer:</u>

The correct answer option is h(f(x)) = 2x - 11

<u>Step-by-step explanation:</u>

We are given two functions:

$f(x) = x-7$ and $h(x) = 2x+3$

and we supposed to find another function which is $h(f(x))$ .

To find $h(f(x))$ , we need to substitute the value of $f(x)$ in the other function i.e. $h(x)$ to get:

$h(f(x)) = h(x-7)$

$h(f(x))= 2(x-7)+3$

$h(f(x))=2x-14+3$

$h(f(x)) 2x-11$

Therefore, the correct answer is h(f(x)) = 2x - 11.

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The width of Florida is 4/5 of its length if the length of Florida is about 450 miles what is the approximate width

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(Q1) Which of the following is true about the function y = 3 • 2x?

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Answer: Third option

Step-by-step explanation:

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$y=3*2^x$

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$y=3*2^{(x+1)}$

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3 years ago

Read 2 more answers

Demand for Tablet Computers The quantity demanded per month, x, of a certain make of tablet computer is related to the average u

soldier1979 [14.2K]

$x = f ( p ) = \frac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } } \\\\ \qquad { p ( t ) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \quad ( 0 \leq t \leq 60 ) }$

Answer:

12.0 tablet computers/month

Step-by-step explanation:

The average price of the tablet 25 months from now will be:

$p ( 25) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { 25 } } + 200 \\= \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \times 5 } + 200\\\\=\dfrac { 400 } { 1 + \dfrac { 5 } { 8 } } + 200\\p(25)=\dfrac { 5800 } {13}$

Next, we determine the rate at which the quantity demanded changes with respect to time.

Using Chain Rule (and a calculator)

$\dfrac{dx}{dt}= \dfrac{dx}{dp}\dfrac{dp}{dt}$

$\dfrac{dx}{dp}= \dfrac{d}{dp}\left[{ \dfrac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } }\right] =-\dfrac{100}{9}p(810,000-p^2)^{-1/2}$

$\dfrac{dp}{dt}=\dfrac{d}{dt}\left[\dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \right]=-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}$

Therefore:

$\dfrac{dx}{dt}= \left[-\dfrac{100}{9}p(810,000-p^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}\right]$

Recall that at t=25, $p(25)=\dfrac { 5800 } {13} \approx 446.15$

Therefore:

$\dfrac{dx}{dt}(25)= \left[-\dfrac{100}{9}\times 446.15(810,000-446.15^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt {25} \right]^{-2}25^{-1/2}\right]\\=12.009$

The quantity demanded per month of the tablet computers will be changing at a rate of 12 tablet computers/month correct to 1 decimal place.

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