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OLga [1]
4 years ago
12

Help if you can please!!

Mathematics
1 answer:
rosijanka [135]4 years ago
6 0
<h3>Answer: Choice (B)</h3><h3>25b^2+40b+23</h3>

===================================================

Work Shown:

c(b) = 5b

h(b) = b+4

h(c(b)) = c(b)+4 ... replace every b with c(b)

h(c(b)) = 5b+4 ... replace c(b) on the right hand side with 5b

For the sake of simplicity, let's call "h(c(b))" as k

k = h(c(b))

k = 5b+4

that way we can use it later in the next section

---------------

a(b) = b^2 + 7

a(k) = k^2 + 7 ... replace b with k

a(k) = (5b+4)^2 + 7 ... plug in k = 5b+4

a(k) = 25b^2+40b+16+7 .... use FOIL rule

a(k) = 25b^2+40b+23

a( h(c(b)) ) = 25b^2+40b+23

In that last step, I replaced k with h(c(b)) so that we get to the original format your teacher wanted.

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49 is the answer.

Step-by-step explanation:

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Answer:  \bold{(D)\ \dfrac{3}{4}-\dfrac{1}{4}x}

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

\ \ \dfrac{3}{4}-x\bigg(\dfrac{1}{2}-\dfrac{5}{8}\bigg)+\bigg(-\dfrac{3}{8}x\bigg)\\\\\\=\dfrac{3}{4}\bigg(\dfrac{2}{2}\bigg)-x\bigg[\dfrac{1}{2}\bigg(\dfrac{4}{4}\bigg)-\dfrac{5}{8}\bigg]+\bigg(-\dfrac{3}{8}x\bigg)\\\\\\=\dfrac{6}{8}-x\bigg(\dfrac{4}{8}-\dfrac{5}{8}\bigg)-\dfrac{3}{8}x\\\\\\=\dfrac{6}{8}-x\bigg(-\dfrac{1}{8}\bigg)-\dfrac{3}{8}x\\\\\\=\dfrac{6}{8}+\dfrac{1}{8}x-\dfrac{3}{8}x\\\\\\=\dfrac{6}{8}-\dfrac{2}{8}x\\\\\\=\dfrac{3}{4}-\dfrac{1}{4}x\quad \text{(reduced both fractions)}

7 0
3 years ago
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Westkost [7]
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3 years ago
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