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3 years ago

I need help!!! Brainliest for anyone who answers correctly on both!!!

Mathematics

1 answer:

timama [110]3 years ago

3 0

Answer:

The first answer: 390m The second answer: x = 7; m<1 = 17; m<2 = 73

Step-by-step explanation:

First problem:

It's asking for the distance he walks by walking around a rectangular box which is the perimeter.

Two of the sides will have length 115m and two will have length 80m. So this perimeter is 115 + 115 + 80 + 80 = 390m.

Second problem:

Complementary angles add to 90 degrees. So m<1 and m<2 added together will equal 90

-4x +45 + 7x + 24 = 90

3x + 69 = 90

3x = 21

x= 7.

Plug back x into both m<1 and m<2 to find the measure of the angles.

m<1 = -4(7) + 45 = 45-28 = 17

m<2 = 7(7) + 24 = 49 + 24 = 73

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3 years ago

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so that the normal vector to $S$ is given by

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$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=-\dfrac{\partial(8-x-2y)}{\partial x}\,\vec\imath-\dfrac{\partial(8-x-2y)}{\partial y}\,\vec\jmath+\vec k=\vec\imath+2\,\vec\jmath+\vec k$

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$\displaystyle\iint_Se^z\,\mathrm d\Sigma=\sqrt6\iint_Te^{8-x-2y}\,\mathrm dy\,\mathrm dx$

where $T$ is the region in the $x,y$ plane over which $S$ is defined. In this case, it's the triangle in the plane $z=0$ which we can capture with $0\le x\le8$ and $0\le y\le\frac{8-x}2$ , so that we have

$\displaystyle\sqrt6\iint_Te^{8-x-2y}\,\mathrm dx\,\mathrm dy=\sqrt6\int_0^8\int_0^{(8-x)/2}e^{8-x-2y}\,\mathrm dy\,\mathrm dx=\boxed{\sqrt{\frac32}(e^8-9)}$

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