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BaLLatris [955]
2 years ago
8

Here is the thing for question two

Mathematics
2 answers:
pogonyaev2 years ago
7 0

Answer:

hey. kiwi account is blocked so... i talk 2 u with this 1 4 a

while? i guess.

Sergio039 [100]2 years ago
6 0

Answer:

26

Step-by-step explanation:

21<x-5

move -5 to other side

21+5<x, 26<x

so answer is 26

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1 pt) If a parametric surface given by r1(u,v)=f(u,v)i+g(u,v)j+h(u,v)k and −4≤u≤4,−4≤v≤4, has surface area equal to 1, what is t
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The area of the surface given by \vec r_1(u,v) is 1. In terms of a surface integral, we have

1=\displaystyle\int_{-4}^4\int_{-4}^4\left\|\frac{\partial\vec r_1(u,v)}{\partial u}\times\frac{\partial\vec r_1(u,v)}{\partial v}\right\|\,\mathrm du\,\mathrm dv

By multiplying each component in \vec r_1 by 5, we have

\dfrac{\partial\vec r_2(u,v)}{\partial u}=5\dfrac{\partial\vec r_1(u,v)}{\partial u}

and the same goes for the derivative with respect to v. Then the area of the surface given by \vec r_2(u,v) is

\displaystyle\int_{-4}^4\int_{-4}^425\left\|\frac{\partial\vec r_1(u,v)}{\partial u}\times\frac{\partial\vec r_1(u,v)}{\partial v}\right\|\,\mathrm du\,\mathrm dv=\boxed{25}

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