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Natali5045456 [20]

2 years ago

15

A pelican is flying 120 feet above sea level. The pelican dives to catch a fish that is 35 feet below sea level. Which number re

presents the distance the pelican will travel?

1 answer:

sergiy2304 [10]2 years ago

8 0

Answer:

155

Step-by-step explanation:

Remember that below sea level means negative. So 120-(-35)=155

I hope this helps! :)

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Correct answers only please!

MrRissso [65]

Answer:

23.6°

Step-by-step explanation:

In this question we have to use some trigonometry to work out angle XVW. Since we are working out angles all of the trigonometric functions will have to be to the ⁻¹. The first thing we need to identify is will formula will we use out of the following:

Sin⁻¹ = Opposite ÷ Hypotenuse

Tan⁻¹ = Opposite ÷ Adjacent

Cos⁻¹ = Adjacent ÷ Hypotenuse

-------------------------------------------------------------------------------------------------------

10 = Hypotenuse

4 = Opposite

-------------------------------------------------------------------------------------------------------

We know which formula to use because the length won't be in the triangle for example we will be using the Sin triangle because we don't have an adjacent. If we don't have an adjacent then the other formula's won't work.

Now we substitute in the values to find the value of angle XVW

Sin⁻¹ = Opposite ÷ Hypotenuse

Sin⁻¹ = 4 ÷ 10

The value of angle XVW is 23.57817848

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2 years ago

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Answer:

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Step-by-step explanation:

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Read 2 more answers

What's the flux of the vector field F(x,y,z) = (e^-y) i - (y) j + (x sinz) k across σ with outward orientation where σ is the po

emmasim [6.3K]

$\displaystyle\iint_\sigma\mathbf F\cdot\mathrm dS$
$\displaystyle\iint_\sigma\mathbf F\cdot\mathbf n\,\mathrm dS$
$\displaystyle\iint_\sigma\mathbf F\cdot\left(\frac{\mathbf r_u\times\mathbf r_v}{\|\mathbf r_u\times\mathbf r_v\|}\right)\|\mathbf r_u\times\mathbf r_v\|\,\mathrm dA$
$\displaystyle\iint_\sigma\mathbf F\cdot(\mathbf r_u\times\mathbf r_v)\,\mathrm dA$

Since you want to find flux in the outward direction, you need to make sure that the normal vector points that way. You have

$\mathbf r_u=\dfrac\partial{\partial u}[2\cos v\,\mathbf i+\sin v\,\mathbf j+u\,\mathbf k]=\mathbf k$
$\mathbf r_v=\dfrac\partial{\partial v}[2\cos v\,\mathbf i+\sin v\,\mathbf j+u\,\mathbf k]=-2\sin v\,\mathbf i+\cos v\,\mathbf j$

The cross product is

$\mathbf r_u\times\mathbf r_v=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\0&0&1\\-2\sin v&\cos v&0\end{vmatrix}=-\cos v\,\mathbf i-2\sin v\,\mathbf j$

So, the flux is given by

$\displaystyle\iint_\sigma(e^{-\sin v}\,\mathbf i-\sin v\,\mathbf j+2\cos v\sin u\,\mathbf k)\cdot(\cos v\,\mathbf i+2\sin v\,\mathbf j)\,\mathrm dA$
$\displaystyle\int_0^5\int_0^{2\pi}(-e^{-\sin v}\cos v+2\sin^2v)\,\mathrm dv\,\mathrm du$
$\displaystyle-5\int_0^{2\pi}e^{-\sin v}\cos v\,\mathrm dv+10\int_0^{2\pi}\sin^2v\,\mathrm dv$
$\displaystyle5\int_0^0e^t\,\mathrm dt+5\int_0^{2\pi}(1-\cos2v)\,\mathrm dv$

where $t=-\sin v$ in the first integral, and the half-angle identity is used in the second. The first integral vanishes, leaving you with

$\displaystyle5\int_0^{2\pi}(1-\cos2v)\,\mathrm dv=5\left(v-\dfrac12\sin2v\right)\bigg|_{v=0}^{v=2\pi}=10\pi$

5 0

3 years ago