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2 years ago

What is the volume of the triangular prism?

Mathematics

2 answers:

DENIUS [597]2 years ago

7 0

Answer:

115.5m ${}^{2}$

Step-by-step explanation:

The volume of a prism = 1/2 a × c × h

according to the image a appears to be 3 m she appears to be 7m. and height appears to be 11m.

=> 1/2 a × c × h

=> 1/2 3m × 7m × 11m

multiply 3, 7 and 11

=> 1/2 231

=> 231/2

=> 115.5m ${}^{2}$

done at all I hope this help please mark me brainest^_^

kogti [31]2 years ago

4 0

Answer:115.5

Step-by-step explanation:

hope this helped!!

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Integrala x la a treia ori ln la a doua dx va rog

Studentka2010 [4]

I don't speak Romanian, but the closest translation for this suggests you're trying to compute

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where

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Integrate by parts again:

$\displaystyle \int x^3 \ln(x) \, dx = u'v' - \int v' du'$

where

u' = ln(x) ⇒ du' = dx/x

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$\implies \displaystyle \int x^3 \ln(x) \, dx = \frac14 x^4 \ln(x) - \frac14 \int x^3 \, dx$

So, we have

$\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac12 \left(\frac14 x^4 \ln(x) - \frac14 \int x^3 \, dx \right)$

$\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac18 x^4 \ln(x) + \frac18 \int x^3 \, dx$

$\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac18 x^4 \ln(x) + \frac18 \left(\frac14 x^4\right) + C$

$\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac18 x^4 \ln(x) + \frac1{32} x^4 + C$

$\boxed{\displaystyle \int x^3 \ln(x)^2 \, dx = \frac1{32} x^4 \left(8\ln(x)^2 - 4\ln(x) + 1\right) + C}$

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