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pickupchik [31]

2 years ago

11

What is the value of x which proves that the runways are parallel? What is the measure of the larger angle? (evidently the answe

r is NOT 'b')
A) x = 20; 92 degrees

B) x = 40; 88 degrees

C) x = 25; 107 degrees

D) x = 30; 115 degrees

1 answer:

romanna [79]2 years ago

3 0

I feel like the answer is C

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Answer the following question. Click on symbol to choose correct answer. Given: R = {(x, y): y = -x^2} What is the range of R? y

3241004551 [841]

Answer: c

I answered c and got a 100 so it has to be it

3 0

2 years ago

HELP!!!!!!<br><br> y = 6kx - 2; m = 2/3

Mrrafil [7]

Answer:

k=1/9

Step-by-step explanation:

6k is the slope , and you actual slope(m) is given, hence,make em equal to each other:

6k = 2/3

k = 2/3(6) [Divide both sides by 6]

k = 1 /9 [Simplify]

Hope this helps!

Mark brainliest if you think I helped!

Would really appreciate!

7 0

2 years ago

Mia buys 3 gallons of gas that cost d dollars per gallon.Bob buys g gallons of gas that cost 3 per gallon. Write an expression f

xxTIMURxx [149]

3(d) and 3 (g)
d represents the price. so you multiply 3 by however much the has cost.
g represents the gallons and 3 represents the cost and again you multiply

3 0

2 years ago

Read 2 more answers

Christy spent $500 over her budget on gifts during the holidays. In addition to her regular job, she took a part-time job at the

Deffense [45]

Christy will have to work 50 hours to pay off her debt.

500 divided by 10 = 50.

Hope this helped!

5 0

2 years ago

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

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

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

###

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":

$\begin{cases}x(t)=4\cos^3t\\y(t)=4\sin^3t\end{cases}$

which gives

$x(t)^{2/3}+y(t)^{2/3}=(4\cos^3t)^{2/3}+(4\sin^3t)^{2/3}=4^{2/3}(\cos^2t+\sin^2t)=4^{2/3}$

as needed. Then with $0\le t\le2\pi$ , we compute the area via Green's theorem using the same setup as before:

$\displaystyle\iint_E\mathrm dx\,\mathrm dy=\frac12\int_0^{2\pi}(-4\sin^3t(12\cos^2t(-\sin t))+4\cos^3t(12\sin^2t\cos t))\,\mathrm dt$

$=\displaystyle24\int_0^{2\pi}(\sin^4t\cos^2t+\cos^4t\sin^2t)\,\mathrm dt$

$=\displaystyle24\int_0^{2\pi}\sin^2t\cos^2t\,\mathrm dt$

$=\displaystyle6\int_0^{2\pi}(1-\cos2t)(1+\cos2t)\,\mathrm dt$

$=\displaystyle6\int_0^{2\pi}(1-\cos^22t)\,\mathrm dt$

$=\displaystyle3\int_0^{2\pi}(1-\cos4t)\,\mathrm dt=6\pi$

3 0

2 years ago