1 year ago

Find h(3)+d(2) if h(x)= x-1 and d(x) = 7x + 3A.2 B.5 C.14 D.19

1 answer:

disa [49]1 year ago

8 0

$\begin{gathered} h(x)=x-1 \\ d(x)=7x+3 \\ h(3)=3-1=2 \\ d(2)=7(2)+3=14+3=17 \\ h(3)+d(2)=2+17=19 \end{gathered}$

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Marina86 [1]

I am assuming this is a triangle.....P = a + b + c

P = 2a - 3 + 2a + 3a + 1....combine like terms
P = 7a - 2 <==

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3 years ago

Reil [10]

Answer: THIRD OPTION.

Step-by-step explanation:

For this exercise you need to use the formula for calculate the distance between two points. This is:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Given the following points:

$(5, 1)$ and $(-3,-5)$

You can identify that:

$x_2=-3\\x_1=5\\\\y_2=-5\\y_1=1$

Knowing these values, the next step is to substitute them into the equation for calculate the distance between two points and then evaluate.

Therefore, the distance between $(5, 1)$ and $(-3,-5)$ is:

$d=\sqrt{(-3-5)^2+(-5-1)^2}\\\\d=10$

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3 years ago

fgiga [73]

Answer:

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3 years ago

IgorLugansk [536]

Answer:

Step-by-step explanation:

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3 years ago

kobusy [5.1K]

(2X-3)(4-3X)
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That's your answer.

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3 years ago