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Y_Kistochka [10]
2 years ago
14

Find the measures of ZX and ZY.

Mathematics
2 answers:
Ulleksa [173]2 years ago
6 0

mzh=hahsvvrbhegdvvbehe

Lelechka [254]2 years ago
5 0

Answer: 303

Step-by-step explanation:

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45 minutes as you only have to divide 60 by 4 and then multiply it times 3
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Integration by Parts Evaluate e-2x cos(2x) dx.​
kifflom [539]

Let

I = \displaystyle \int e^{-2x} \cos(2x) \, dx[/]texIntegrate by parts:[tex]\displaystyle \int u \, dv = uv - \int v \, du

with

u = e^{-2x} \implies du = -2 e^{-2x} \, dx \\\\ dv = \cos(2x) \, dx \implies v = \dfrac12 \sin(2x)

Then

\displaystyle I = \frac12 e^{-2x} \sin(2x) + \int e^{-2x} \sin(2x) \, dx + C

Integrate by parts again, this time with

u = e^{-2x} \implies du = -2 e^{-2x} \, dx \\\\ dv = \sin(2x) \, dx \implies v = -\dfrac12 \cos(2x)

so that

\displaystyle I = \frac12 e^{-2x} \sin(2x) - \frac12 e^{-2x} \cos(2x) - \int e^{-2x} \cos(2x) \, dx + C\\\\ \implies I = \frac{\sin(2x)-\cos(2x)}{2e^{2x}} - I + C \\\\ \implies 2I = \frac{\sin(2x) - \cos(2x)}{2e^{2x}} + C \\\\ \implies I = \boxed{\frac{\sin(2x) - \cos(2x)}{4e^{2x}} + C}

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1 year ago
Mr. Ross is purchasing a table and chairs for
GarryVolchara [31]
B. r = 75m + 1350 I think is the answer. I hope this helped you!
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3 years ago
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Vlad [161]

A. 5+x=17, x = 17-5=12

B. x = 4 x 2.5 = 10

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3 0
2 years ago
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