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2 years ago

Find the surface area of the regular pyramid

Mathematics

1 answer:

kicyunya [14]2 years ago

5 0

Pyramid Surface Area = (½ * Perimeter of Base * Slant Height) + Base Area

Pyramid Surface Area = (.5 * 56 * 11) + 14 * 14Pyramid Surface Area = 308 + 196Pyramid Surface Area =504 square millimeters

Source:http://www.1728.org/volpyrmd.htm

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Use the given instructions to answer question 18 to question 20.

Tpy6a [65]

Answer:

Explanation:

6 0

7 months ago

Consider F and C below. F(x, y, z) = y2 sin(z) i + 2xy sin(z) j + xy2 cos(z) k C: r(t) = t2 i + sin(t) j + t k, 0 ≤ t ≤ π (a) Fi

Liono4ka [1.6K]

Answer:

a) $f (x,y,z)= xy^2\sin(z)$

b) $\int_C F \cdot dr =0$

Step-by-step explanation:

Recall that given a function f(x,y,z) then $\nabla f = (\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z})$ . To find f, we will assume it exists and then we will find its form by integration.

First assume that F = $\nabla f$ . This implies that

$\frac{\partial f}{\partial x} = y^2\sin(z)$ if we integrate with respect to x we get that

$f(x,y,z) = xy^2\sin(z) + g(y,z)$ for some function g(y,z). If we take the derivative of this equation with respect to y, we get

$\frac{\partial f}{\partial y} = 2xy\sin(z) + \frac{\partial g}{\partial y}$

This must be equal to the second component of F. Then

$2xy\sin(z) + \frac{\partial g}{\partial y}=2xy\sin(z)$

This implies that $\frac{\partial g}{\partial y}=0$ , which means that g depends on z only. So $f(x,y,z) = xy^2\sin(z) + g(z)$

Taking the derivative with respect to z and making it equal to the third component of F, we get

$xy^2\cos(z)+\frac{dg}{dz} = xy^2\cos(z)$

which implies that $\frac{dg}{dz}=0$ which means that g(z) = K, where K is a constant. So

$f (x,y,z)= xy^2\sin(z)$

b) To evaluate $\int_C F \cdot dr$ we can evaluate it by using f. We can calculate the value of f at the initial and final point of C and the subtract them as follows.

$\int_C F \cdot dr = f(r(\pi))-f(r(0))$

Recall that $r(\pi) = (\pi^2, 0, \pi)$ so $f(r(\pi)) = \pi^2\cdot 0 \cdot \sin(\pi) = 0$

Also $r(0) = (0, 0, 0)$ so $f(r(0)) = 0^2\cdot 0 \cdot \sin(0) = 0$

So $\int_C F \cdot dr =0$

5 0

2 years ago

the height, h cm, of a table is measured s 72cm correct to the nearest centimeters. conplete the following statement to show the

gogolik [260]

Anything between 71.5 and 72.499 recurring.

Hope this helps :)

6 0

2 years ago

Please help me it’s due today at 11:59

lesya [120]

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