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3 years ago

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The triangles are similar. If QR = 9, QP = 6, and TU = 19, find TS. Round to the nearest tenth.

1 answer:

Feliz [49]3 years ago

4 0

QR=9
QP=6
TU=19
TS=?
Okay so the triangles are kindred, now the equation can be made
9/16=6/x
So now we are going to cross multiply
9x=114
Now divide both sides by
912 2/3
Your answer is:
TS=12.6

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Step-by-step explanation:

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Using sum rule of integrals, we will get:

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$\int\ vdv=uv-\int\ vdu$

Let $u=x$ and $v'=\frac{1}{e^x}$ .

Now, we need to find du and v using these values as shown below:

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$v=-\frac{1}{e^x}$

Substituting our given values in integration by parts formula, we will get:

$\frac{1}{3}\int _0^3\frac{x}{e^x}dx=\frac{1}{3}(x*(-\frac{1}{e^x})-\int _0^3(-\frac{1}{e^x})dx)$

$\frac{1}{3}\int _0^3\frac{x}{e^x}dx=\frac{1}{3}(-\frac{x}{e^x}- (\frac{1}{e^x}))$

$\int _0^3\:3dx-\int _0^3\frac{x}{3e^x}dx=3x-\frac{1}{3}(-\frac{x}{e^x}- (\frac{1}{e^x}))$

Compute the boundaries:

$3(3)-\frac{1}{3}(-\frac{3}{e^3}- (\frac{1}{e^3}))=9+\frac{4}{3e^3}=9.06638$

$3(0)-\frac{1}{3}(-\frac{0}{e^0}- (\frac{1}{e^0}))=0-(-\frac{1}{3})=\frac{1}{3}$

$9.06638-\frac{1}{3}=8.733046$

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