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3 years ago

7

Plz help question below

2 answers:

lesya692 [45]3 years ago

6 0

I believe it is a. hope this helps!!!!!!!

Olenka [21]3 years ago

3 0

It should be A if I did the math right which I think I did

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Two questions in picture! Need help!

kkurt [141]

Answer:

1) C) sin(θ) = 119/169

2) D) cos(θ) = 120/169

Step-by-step explanation:

The mnemonic SOH CAH TOA expresses the relationships you need for answering these questions.

Sin(θ) = Opposite/Hypotenuse = 119/169 . . . . . problem 1

Cos(θ) = Adjacent/Hypotenuse = 120/169 . . . . problem 2

5 0

2 years ago

Use Stokes' Theorem to evaluate C F · dr F(x, y, z) = xyi + yzj + zxk, C is the boundary of the part of the paraboloid z = 1 − x

Serggg [28]

I assume $C$ has counterclockwise orientation when viewed from above.

By Stokes' theorem,

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

so we first compute the curl:

$\vec F(x,y,z)=xy\,\vec\imath+yz\,\vec\jmath+xz\,\vec k$

$\implies\nabla\times\vec F(x,y,z)=-y\,\vec\imath-z\,\vec\jmath-x\,\vec k$

Then parameterize $S$ by

$\vec r(u,v)=\cos u\sin v\,\vec\imath+\sin u\sin v\,\vec\jmath+\cos^2v\,\vec k$

where the $z$ -component is obtained from

$1-(\cos u\sin v)^2-(\sin u\sin v)^2=1-\sin^2v=\cos^2v$

with $0\le u\le\dfrac\pi2$ and $0\le v\le\dfrac\pi2$ .

Take the normal vector to $S$ to be

$\vec r_v\times\vec r_u=2\cos u\cos v\sin^2v\,\vec\imath+\sin u\sin v\sin(2v)\,\vec\jmath+\cos v\sin v\,\vec k$

Then the line integral is equal in value to the surface integral,

$\displaystyle\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

$=\displaystyle\int_0^{\pi/2}\int_0^{\pi/2}(-\sin u\sin v\,\vec\imath-\cos^2v\,\vec\jmath-\cos u\sin v\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{\pi/2}\int_0^{\pi/2}\cos v\sin^2v(\cos u+2\cos^2v\sin u+\sin(2u)\sin v)\,\mathrm du\,\mathrm dv=\boxed{-\frac{17}{20}}$

6 0

3 years ago

#12 with explanation please

4vir4ik [10]

Answer:

30.01 units

Step-by-step explanation:

<u>Given vertices WXYZ:</u>

W(-4,5), X(2,4), Y(3, -4), Z(-1, -6)

Perimeter is the sum of the side length.

<u>Use distance formula to find each side:</u>

WX = $\sqrt{(2+4)^2+(4-5)^2} =\sqrt{37}$ ≈ 6.08
XY = $\sqrt{(3-2)^2+(-4-4)^2} =\sqrt{65}$ ≈ 8.06
YZ = $\sqrt{(-1-3)^2+(-6+4)^2} =\sqrt{20}$ ≈ 4.47
WZ = $\sqrt{(-1+4)^2+(-6-5)^2} =\sqrt{130}$ $\sqrt{(-1+4)^2+(-6-5)^2}=\sqrt{130}$ ≈ 11.40

<u>Find perimeter:</u>

P = 6.08 + 8.06 + 4.47 + 11.40 = 30.01 units

7 0

3 years ago

What number is 80% of 90%?

storchak [24]

Answering so you can mark the other person brainliest

6 0

2 years ago

What ordered pair is a solution of the equation?<br><br> y=-2x-5

ioda

$\sf \underline{Answer:}$

Pick any value for x, like x = 0, plug it into the equation, and get the value for y. For x = 0,

2(0) + y = 5

y = 5

The ordered pair is (0,5). Find other ordered pairs using the same reasoning.

3 0

1 year ago

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