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Aleks [24]
3 years ago
13

square A has a side length (2x-7) and square B has side length (-4x+18) how much bigger is the perimeter of square B than square

A?
Mathematics
1 answer:
Paladinen [302]3 years ago
4 0
To make things easier, we can assume a value of x in order to substitute to each expression and obtain the side length of each. We can calculate for perimeter of each square. Then, compare them. We do as follows:

Assume x = 4
Square A
2x - 7 = 2(4) - 7 = 1
P = 4(1) = 4

Square B
-4x + 18 = -4(4) + 18 = 2
P = 4(2) = 8

Therefore, the perimeter of B is 2 than the perimeter of A.

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Leya [2.2K]
Parameterize the lateral face T_1 of the cylinder by

\mathbf r_1(u,v)=(x(u,v),y(u,v),z(u,v))=(2\cos u,2\sin u,v

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\mathbf r_2(r,\theta)=(x(r,\theta),y(r,\theta),z(r,\theta))=(r\cos\theta,r\sin\theta,0)
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The integral along the surface of the cylinder (with outward/positive orientation) is then

\displaystyle\iint_S(x^2+y^2+z^2)\,\mathrm dS=\left\{\iint_{T_1}+\iint_{T_2}+\iint_{T_3}\right\}(x^2+y^2+z^2)\,\mathrm dS
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=\displaystyle2\int_{u=0}^{u=2\pi}\int_{v=0}^{v=3}(v^2+4)\,\mathrm dv\,\mathrm du+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}r^3\,\mathrm d\theta\,\mathrm dr+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}r(r^2+9)\,\mathrm d\theta\,\mathrm dr
=\displaystyle4\pi\int_{v=0}^{v=3}(v^2+4)\,\mathrm dv+2\pi\int_{r=0}^{r=2}r^3\,\mathrm dr+2\pi\int_{r=0}^{r=2}r(r^2+9)\,\mathrm dr
=136\pi
7 0
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Answer:

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add them together you will get -21

6 0
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