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brilliants [131]

2 years ago

15

What is the midpoint of of VX, find the length of VX

1 answer:

Rashid [163]2 years ago

8 0

The midpoint is exactly in the middle
meaning
VW = WX
3x+7 = x + 25
solve from x
3x + 7 = x + 25
subtract 7 from both sides
3x = x + 18
subtract x from both sides
2x = 18
divide both sides by 2
x = 9
Now we have x = 9 but we aren't done.

plug what we found into the expressions
3x + 7
3(9) + 7
27 + 7
34

x + 25
9 + 25
34

34 + 34
68

The length of VX is 68.
Or option D.) 68

Hope this helps

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Read 2 more answers

Consider the function g(x) = (x-e)^3e^-(x-e). Find all critical points and points of inflection (x, g(x)) of the function g.

Elden [556K]

Answer:

The answer is " $cirtical\ points \ (x,g(x))\equiv (e,0),(e+3,\frac{27}{e^3})$ "

Step-by-step explanation:

Given:

$g(x) = (x-e)^3e^{-(x-e)}$

Find critical points:

$g(x) = (x-e)^3e^{(e-x)}$

differentiate the value with respect of x:

$\to g'(x)= (x-e)^3 \frac{d}{dx}e^{e-r} +e^{e-r} \frac{d}{dx}(x-e)^3=(x-e)^2 e^{(e-x)} [-x+e+3]$

critical points $g'(x)=0$

$\to (x-e)^2 e^{(e-x)} [e+3-x]=0\\\\\to e^{(e-x)}\neq 0 \\\\\to (x-e)^2=0\\\\ \to [e+3-x]=0\\\\\to x=e\\\\\to x=e+3\\\\\to x= e,e+3$

So,

The critical points of $(x,g(x))\equiv (e,0),(e+3,\frac{27}{e^3})$

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