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3 years ago

Ella calculated the missing side length of one of these triangles using the Pythagorean Theorem. Which triangle is it?

Mathematics

1 answer:

torisob [31]3 years ago

3 0

I think that it is a

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Solve this parallelogram angle

Vesnalui [34]

Answer:

103

Step-by-step explanation:

angles formed on the same line of a parallelogram are supplementary angles

( the angle with the measure of 77 degrees and the angle labeled "x" are formed on the same line which means they are supplementary )

supplementary angles add up to equal 180

hence, 77 + x = 180

180 - 77 = 103

Hence, x = 103

7 0

2 years ago

Read 2 more answers

What function equation is represented by the graph?

dezoksy [38]

Answer:

maybe f(x)= -2/3 x + 1 since it goes down two times or up two times and three times to the left or right depending where you wanna go i don't know how to explain but maybe this might help

6 0

3 years ago

Solve the folloeing equation for a number x. 5x-x-6=6(2x+7) plllllzzz help

Ray Of Light [21]

Answer:

x=-6

Step-by-step explanation:

5x-x-6=6(2x+7)

4x-6=6(2x+7)

4x-6=12x+42

-8x=48 |:(-8)

x=-6

3 0

2 years ago

A class of 40 students elected a class president. There were 12 votes for Candidate A, and 18 votes for Candidate B. The remaini

scoundrel [369]

Hey You!

12/40 = Candidate A.

18/40 = Candidate B.

12 + 18 = 30

40 - 30 = 10

10/40 = Those who didn't vote.

ANSWER:

12 ÷ 40 = .3 (30%) So 30% voted for Candidate A.

18 ÷ 40 = .45 (45%) So 45% voted for Candidate B.

10 ÷ 40 = .25 (25%) So 25% didn't vote at all.

Those Are Your Answers! ^^^

3 0

3 years ago

Read 2 more answers

Use green's theorem to compute the area inside the ellipse x252+y2172=1. use the fact that the area can be written as ∬ddxdy=12∫

Pavel [41]

The area of the ellipse $E$ is given by

$\displaystyle\iint_E\mathrm dA=\iint_E\mathrm dx\,\mathrm dy$

To use Green's theorem, which says

$\displaystyle\int_{\partial E}L\,\mathrm dx+M\,\mathrm dy=\iint_E\left(\frac{\partial M}{\partial x}-\frac{\partial L}{\partial y}\right)\,\mathrm dx\,\mathrm dy$

( $\partial E$ denotes the boundary of $E$ ), we want to find $M(x,y)$ and $L(x,y)$ such that

$\dfrac{\partial M}{\partial x}-\dfrac{\partial L}{\partial y}=1$

and then we would simply compute the line integral. As the hint suggests, we can pick

$\begin{cases}M(x,y)=\dfrac x2\\\\L(x,y)=-\dfrac y2\end{cases}\implies\begin{cases}\dfrac{\partial M}{\partial x}=\dfrac12\\\\\dfrac{\partial L}{\partial y}=-\dfrac12\end{cases}\implies\dfrac{\partial M}{\partial x}-\dfrac{\partial L}{\partial y}=1$

The line integral is then

$\displaystyle\frac12\int_{\partial E}-y\,\mathrm dx+x\,\mathrm dy$

We parameterize the boundary by

$\begin{cases}x(t)=5\cos t\\y(t)=17\sin t\end{cases}$

with $0\le t\le2\pi$ . Then the integral is

$\displaystyle\frac12\int_0^{2\pi}(-17\sin t(-5\sin t)+5\cos t(17\cos t))\,\mathrm dt$

$=\displaystyle\frac{85}2\int_0^{2\pi}\sin^2t+\cos^2t\,\mathrm dt=\frac{85}2\int_0^{2\pi}\mathrm dt=85\pi$

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Notice that $x^{2/3}+y^{2/3}=4^{2/3}$ kind of resembles the equation for a circle with radius 4, $x^2+y^2=4^2$ . We can change coordinates to what you might call "pseudo-polar":