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3 years ago

20 PTS!!! Math pls help

Mathematics

1 answer:

Rainbow [258]3 years ago

3 0

Answer:

The answer to your question is: 30 u²

Step-by-step explanation:

Process

1.- Calculate the length of the base and the length of the height

dAB = $\sqrt{(x2-x1)^{2} + (y2 - y1)^{2} }$

dAB = $\sqrt{(5+ 3)^{2} + (1 - 5)^{2} }$

dAB = $\sqrt{8^{2} + (-4)^{2} }$

dAB = $\sqrt{64 + 16}$

dAB = $\sqrt{80}$

dCD = $\sqrt{(x2-x1)^{2} + (y2 - y1)^{2} }$

dCD = $\sqrt{(-4 + 1)^{2} + (-2 - 4)^{2} }$

dCD = $\sqrt{(-3)^{2} + (-6)^{2} }$

dCD = $\sqrt{9 + 36}$

dCD = $\sqrt{45}$

Area of a triangle = base x height / 2

= $\frac{\sqrt{80} \sqrt{45} }{2}$

$= \frac{\sqrt{3600} }{2}$

$= \frac{60}{2}$

= 30 u²

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Can anyone please help, ;_;

BlackZzzverrR [31]

It was hard to figure this put but I'll just a take an educated guess and say that the answer is 10%.

7 0

3 years ago

Can you help me solve this?

antiseptic1488 [7]

The required composite value f(g(-8) is 28

A composite function is a function inside another function. Given the following functions

f(x) = x² - 3x - 12

g(x) = -x - 12

Required composite function is f(g(-8))

First we need to get f(g(x))

f(g(x)) = f(-x-12)

f(-x-12) = (-x-12)² - 3x - 12

Expand

f(-x-12) = x² +24x + 144 - 3x - 12

f(-x-12) = x² +21x + 132

f(g(x)) = x² +21x + 132

Substitute x = -8 into the result

f(g(-8)) = (-8)² +21(-8) + 132

f(g(-8)) = 64 - 168 + 132

f(g(-8)) = 28

Hence the required composite value f(g(-8) is 28

Learn more here : brainly.com/question/3256461

8 0

3 years ago

Let M be the closed surface that consists of the hemisphere

ycow [4]

Since $M$ is closed, you can use the divergence theorem: The flux of $\vec E(x,y,z)$ across $M$ is

$\displaystyle\iint_{\partial M}\vec E\cdot\mathrm d\vec S=\iiint_M(\nabla\cdot\vec E)\,\mathrm dV=54\iiint_M\mathrm dV$

which is 54 times the volume of the hemisphere centered at (0, 0, 0) with radius 1, $\boxed{36\pi}$ .

Judging by the question content, you're supposed to find this value by computing the the integral of $\vec E$ across $M_1$ and $M_2$ .

Across $M_1$ :

Parameterize the hemisphere by

$\vec r(u,v)=(\cos u\sin v,\sin u\sin v,\cos v)$

with $0\le u\le2\pi$ and $0\le v\le\frac\pi2$ . Take the normal vector to $M_1$ to be

$\dfrac{\partial\vec r}{\partial v}\times\dfrac{\partial\vec r}{\partial u}=(\cos u\sin^2v,\sin u\sin^2v,\sin v\cos v)$

The flux of $\vec E$ across $M_1$ is

$\displaystyle\iint_{M_1}\vec E\cdot\mathrm d\vec S=18\int_0^{\pi/2}\int_0^{2\pi}(\cos u\sin v,\sin u\sin v,\cos v)\cdot\left(\dfrac{\partial\vec r}{\partial v}\times\dfrac{\partial\vec r}{\partial u}\right)\,\mathrm du\,\mathrm dv$

$=\displaystyle18\int_0^{\pi/2}\int_0^{2\pi}\sin v\,\mathrm du\,\mathrm dv=36\pi$

Across $M_2$ :

Parameterize the disk by

$\vec s(u,v)=(u\cos v,u\sin v,0)$

with $0\le u\le1$ and $0\le v\le2\pi$ . Take the normal to $M_2$ to be

$\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec u}{\partial v}=(0,0,-u)$

Then the flux across $M_2$ is

$\displaystyle\iint_{M_2}\vec E\cdot\mathrm d\vec S=18\int_0^{2\pi}\int_0^1(u\cos v,u\sin v,0)\cdot\left(\frac{\partial\vec s}{\partial v}\times\frac{\partial\vec s}{\partial u}\right)\,\mathrm du\,\mathrm dv=0$