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Nonamiya [84]
3 years ago
11

Hey!

Mathematics
1 answer:
noname [10]3 years ago
5 0

Answer:

A. Cylinder + cone

<u>Volume is the sum of volumes:</u>

  • V = Vcon + Vcyl = 1/3πr²h₁ + πr²h₂
  • V = 1/3π*9²*12 + π*9²*120 = 31554.2 cm³

<u>Surface area of cone:</u>

  • A = A=πr(r+√(h₁²+r²)) = π*9(9 + √(9²+12²)) = 678.6 cm²

<u>Surface area of cylinder minus bases:</u>

  • A = 2πrh₂ = 2π*9*120 = 6785.8 cm²

<u>Total surface area:</u>

  • 678.6 + 6785.8 = 7464.4 cm²

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

B. Cube+ pyramid

<u>Volume:</u>

  • V = a³ + (1/3)a²h = a³ + (1/3)a²√(l²-(a/2)²)
  • V = 8³ + (1/3)8²√(10²-4²) = 707.5 cm³

<u>Surface area of pyramid:</u>

  • A = a² + 2al = 8² + 2*8*10 = 224 cm²

<u>Surface area of cube minus bases:</u>

  • A = 4a² = 4(8²) = 256 cm²

<u>Total surface area:</u>

  • A = 224 + 256 = 480 cm²
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<img src="https://tex.z-dn.net/?f=5%28sin%28t%29%20%20%5Cfrac%7Bdy%7D%7Bdx%7D%20%2Bycos%28t%29%29%29%3Dcos%28t%29%20%20%28sin%28
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Assuming you mean

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This ODE is linear in y, and you can already contract the left hand side as the derivative of a product:

\dfrac{\mathrm d}{\mathrm dt}\left[5y\sin t\right]=\cos t\sin^2t

Integrating both sides with respect to t yields

5y\sin t=\displaystyle\int\cos t\sin^2t\,\mathrm dt
5y\sin t=\dfrac13\sin^3t+C
y=\dfrac1{15}\sin^2t+C\csc t

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9=\dfrac1{15}\sin^2\dfrac\pi2+C\csc\dfrac\pi2
9=\dfrac1{15}+C
C=\dfrac{134}{15}

so that the particular solution over the interval is

y=\dfrac1{15}\sin^2t+\dfrac{134}{15}\csc t
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