3(2−0.9h)+(−1.3h−4) Please help!!!!!!!!!!!!!!!!!

Mathematics

2 answers:

murzikaleks [220]3 years ago

6 0

3⋅2−3⋅0.9h+1.3h−4⇒

6−2.7h+1.3h−4⇒

6−4−2.7h+1.3h⇒

2−1.4h

This is the answer.

wlad13 [49]3 years ago

6 0

I believe the answer is -2+-4h.
3(2+-0.9h)+(-1.3h+-4)
6+-2.7h+-1.3h+-4
2+-4h
This is what I got

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F ind the volume under the paraboloid z=9(x2+y2) above the triangle on xy-plane enclosed by the lines x=0, y=2, y=x

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Answer:

The answer is 48 units³

Step-by-step explanation:

If we simply draw out the region on the x-y plane enclosed between these lines we realize that,if we evaluate the integral the limits all in all cannot be constants since one side of the triangular region is slanted whose equation is given by y=x. So the one of the limit of one of the integrals in the double integral we need to evaluate must be a variable. We choose x part of the integral to have a variable limit, we could well have chosen y's limits as non constant, but it wouldn't make any difference. So the double integral we need to evaluate is given by,

$V=\int\limits^2_0 {} \, \int\limits^{x=y}_0 {z} \, dx dy\\V=\int\limits^2_0 {} \, \int\limits^{x=y}_0 {9(x^{2}+y^{2})} \, dx dy$

Please note that the order of integration is very important here.We cannot evaluate an integral with variable limit last, we have to evaluate it first.after performing the elementary x integral we get,

$V=9\int\limits^2_0 {4y^{3}/3} \, dy$