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3 years ago

Plsssss helps 50 points

Mathematics

2 answers:

PtichkaEL [24]3 years ago

7 0

Answer:

2000.2 cm³

Step-by-step explanation:

Finding the volume of a cylinder is quite simple if you under what pi (3.14) is. You can watch khan academy. It has some useful videos on solving for volume. It also explains very well. Hope this helps!

lutik1710 [3]3 years ago

3 0

Here's the solution,

volume of cylinder :

=》

$\pi r {}^{2} h$

where,

r = radius
h = height

now, let's solve

=》

$3.14 \times 7 \times 7 \times 13$

=》

$2000.18$

So, volume = 2000.2 cm³

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3 years ago

What would x equal to if x equaled 4 in this: 4x

andriy [413]

Answer:

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Step-by-step explanation:

For this problem, we will simply look at the wording of the question and the given expression. The question ask, "What would x be equal to if x equaled 4 in this: 4x?"

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3 years ago

The area of the triangle below is 12 square inches. What is the length of the base?

Aloiza [94]

Answer:

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Step-by-step explanation:

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4 0

3 years ago

Read 2 more answers

If c is the line segment connecting (x1,y1) to (x2,y2), show that the line integral of xdy-ydx=x1y2-x2y1......this is in a chapt

MatroZZZ [7]

Green's theorem doesn't really apply here. GT relates the line integral over some *closed* connected contour that bounds some region (like a circular path that serves as the boundary to a disk). A line segment doesn't form a region since it's completely one-dimensional.

At any rate, we can still compute the line integral just fine. It's just that GT is irrelevant.

We parameterize the line segment by

$\mathbf r(t)=\langle x_1,y_1\rangle(1-t)+\langle x_2,y_2\rangle t$
$\implies\mathbf r(t)=\langle x_1+(x_2-x_1)t,y_1+(y_2-y_1)t\rangle$

with $0\le t\le1$ . Then we find the differential:

$\mathbf r(t)\equiv\langle x,y\rangle=\langle x_1+(x_2-x_1)t,y_1+(y_2-y_1)t\rangle$
$\implies\mathrm d\mathbf r\equiv\langle\mathrm dx,\mathrm dy\rangle=\langle x_2-x_1,y_2-y_1\rangle\,\mathrm dt$

with $0\le t\le1$ .

Here, the line integral is

$\displaystyle\int_{\mathcal C}x\,\mathrm dy-y\,\mathrm dx=\int_{\mathcal C}\langle-y,x\rangle\cdot\langle\mathrm dx,\mathrm dy\rangle$
$=\displaystyle\int_{t=0}^{t=1}\langle-y_1-(y_2-y_1)t,x_1+(x_2-x_1)t\rangle\cdot\langle x_2-x_1,y_2-y_1\rangle\,\mathrm dt$
$=\displaystyle\int_{t=0}^{t=1}(x_1y_2-x_2y_1)\,\mathrm dt$
$=(x_1y_2-x_2y_1)\displaystyle\int_{t=0}^{t=1}\,\mathrm dt$
$=x_1y_2-x_2y_1$

as required.

4 0

4 years ago