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const2013 [10]
2 years ago
8

Find the missing length indicated. Pls show ur work.

Mathematics
2 answers:
Mila [183]2 years ago
7 0

Answer:

33 -  - 9 = 24. \\  \frac{24}{9}  =  \frac{40}{ - x + 30}  \\ 24( - x + 30) = 320 \\  - 24x + 720 = 320 \\  - 24x = 320 - 720 =  - 400 \\ x =  \frac{ - 400}{ - 24}  \\ x = 16.6.

artcher [175]2 years ago
3 0

33-9=24

\\ \sf\longmapsto \dfrac{24}{9}=\dfrac{40}{-x+30}

\\ \sf\longmapsto 24(-x+30)=320

\\ \sf\longmapsto -24x+720=320

\\ \sf\longmapsto -24x=320-720=-400

\\ \sf\longmapsto x=\dfrac{-400}{-24}

\\ \sf\longmapsto x=16.6

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Read 2 more answers
I’m Really lost if I could get a answer it would be greatly appreciated
Nataly [62]
<h3>Answers:</h3>

f(g(x)) = \sqrt{x^2+5}+5\\\\g(f(x)) = x+30+10\sqrt{x-1}

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

Work Shown:

Part 1

f(x) = \sqrt{x-1}+5\\\\f(g(x)) = \sqrt{g(x)-1}+5\\\\f(g(x)) = \sqrt{x^2+6-1}+5\\\\f(g(x)) = \sqrt{x^2+5}+5\\\\

Notice how I replaced every x with g(x) in step 2. Then I plugged in g(x) = x^2+6 and simplified.

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

Part 2

g(x) = x^2+6\\\\g(f(x)) = \left(f(x)\right)^2+6\\\\g(f(x)) = \left(\sqrt{x-1}+5\right)^2+6\\\\g(f(x)) = \left(\sqrt{x-1}\right)^2+2*5*\sqrt{x-1}+\left(5\right)^2+6\\\\g(f(x)) = x-1+10\sqrt{x-1}+25+6\\\\g(f(x)) = x+30+10\sqrt{x-1}\\\\

In step 4, I used the rule (a+b)^2 = a^2+2ab+b^2

In this case, a = sqrt(x-1) and b = 5.

You could also use the box method as a visual way to expand out \left(\sqrt{x-1}+5\right)^2

6 0
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Hope this helped! Comment if you have questions!







8 0
3 years ago
HELP<br> what is the value of “ c”
Alex

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Step-by-step explanation:

When you match the form

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to the equation

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it is reasonable to conclude that ...

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_____

The question presumes you are familiar with the form

... ax² +bx +c = 0

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