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2 years ago

Find the area of the figure shown. Can someone help me please

Mathematics

2 answers:

Lina20 [59]2 years ago

8 0

Answer:

180 cm^2

Step-by-step explanation:

(8 x 15)/2 = 60 cm^2

(16 x 15)/ 2 = 120 cm^2

120 + 60 = 180 cm^2

lisabon 2012 [21]2 years ago

6 0

180 cm2....

..........

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The area of a rectangular glass slab is 80 square centimeters and its perimeter is 36 centimeters. If the length and width of th

netineya [11]

First, you have to find out the values of the length and width.

Let l = length and w = width.

Set up your equation for area: l * w = 80

Set up your equation for perimeter: 2l + 2w = 36

We can use substitution for this problem, so find out the value of l or w. In this case, I will use the first equation to find out the value of l.

(Divide w on both sides) l = 80 / w

Since I know that l = 80 / w, I can substitute it into the second equation.

2l + 2w = 36, and l = 80 / w

= 2 (80/w) + 2w = 36

Simplify the equation.

= 160/w + 2w = 36

Get rid of the w in the denominator by multiplying the entire equation by w.

= ( 160/w + 2w = 36 ) * w

= 160 + $2w^{2}$ = 36w

= 2 ( w - 8 ) ( w - 10 )

The solutions for width (w) is 8 or 10. Then, you can use both 8 and 10 as your length and width, either or is fine. Multiply by 4 for each of those values and you get 8 x 4 = 32 and 10 x 4 = 40. The new perimeter would be 2(32) + 2(40) =

a. 144 cm

Hope this helped you! Let me know if there are any confusions :)

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3 years ago

Let f(x,y,z) = ztan-1(y2) i + z3ln(x2 + 1) j + z k. find the flux of f across the part of the paraboloid x2 + y2 + z = 3 that li

Sophie [7]

Consider the closed region $V$ bounded simultaneously by the paraboloid and plane, jointly denoted $S$ . By the divergence theorem,

$\displaystyle\iint_S\mathbf f(x,y,z)\cdot\mathrm dS=\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV$

And since we have

$\nabla\cdot\mathbf f(x,y,z)=1$

the volume integral will be much easier to compute. Converting to cylindrical coordinates, we have

$\displaystyle\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV=\iiint_V\mathrm dV$
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$=\dfrac\pi2$

Then the integral over the paraboloid would be the difference of the integral over the total surface and the integral over the disk. Denoting the disk by $D$ , we have

$\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-\iint_D\mathbf f\cdot\mathrm dS$

Parameterize $D$ by

$\mathbf s(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf j+2\,\mathbf k$
$\implies\mathbf s_u\times\mathbf s_v=u\,\mathbf k$

which would give a unit normal vector of $\mathbf k$ . However, the divergence theorem requires that the closed surface $S$ be oriented with outward-pointing normal vectors, which means we should instead use $\mathbf s_v\times\mathbf s_u=-u\,\mathbf k$ .

Now,

$\displaystyle\iint_D\mathbf f\cdot\mathrm dS=\int_{u=0}^{u=1}\int_{v=0}^{v=2\pi}\mathbf f(x(u,v),y(u,v),z(u,v))\cdot(-u\,\mathbf k)\,\mathrm dv\,\mathrm du$
$=\displaystyle-4\pi\int_{u=0}^{u=1}u\,\mathrm du$
$=-2\pi$

So, the flux over the paraboloid alone is

$\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-(-2\pi)=\dfrac{5\pi}2$

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