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3 years ago

8

Help plz due today i dont want to fail!

1 answer:

4vir4ik [10]3 years ago

4 0

I believe it’s 1/2 hope it helps

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If c is the curve given by \mathbf{r} \left( t \right = \left( 1 5 \sin t \right \mathbf{i} \left( 1 3 \sin^{2} t \right \mathbf

jonny [76]

With the curve $C$ parameterized by

$C:\mathbf r(t)=\underbrace{15\sin t}_{x(t)}\,\mathbf i+\underbrace{13\sin^2t}_{y(t)}\,\mathbf j+\underbrace{12\sin^3t}_{z(t)}\,\mathbf k$

with $0\le t\le\dfrac\pi2$ , and given the vector field

$\mathbf f(x,y,z)=x\,\mathbf i+y\,\mathbf j+z\,\mathbf k$

the work done by $\mathbf f$ on a particle moving on along $C$ is given by the line integral

$\displaystyle\int_C\mathbf f\cdot\mathrm d\mathbf r=\int\limits_{t=0}^{t=\pi/2}\mathbf f(x(t),y(t),z(t))\cdot\frac{\mathrm d\mathbf r(t)}{\mathrm dt}\,\mathrm dt$

where

$\mathrm d\mathbf r=(15\cos t\,\mathbf i+26\sin t\cos t\,\mathbf j+36\sin^2t\cos t\,\mathbf k)\,\mathrm dt$

The integral is then

$\displaystyle\int_0^{\pi/2}(15\sin t\,\mathbf i+13\sin^2t\,\mathbf j+12\sin^3t\,\mathbf k)\cdot(15\cos t\,\mathbf i+13\sin2t\,\mathbf j+18\sin t\sin2t\,\mathbf k)\,\mathrm dt$
$=\displaystyle\int_0^{\pi/2}(432\sin^5t\cos t+338\sin^3t\cos t+225\sin t\cos t$
$=269$

6 0

4 years ago

Help me please due in 2 minutes

frez [133]

7:4 is the constant and is your answer

5 0

3 years ago

Read 2 more answers

I need help! Will mark Brainliest for full answer!!!

Softa [21]

<u><em>Answer:</em></u>

Part a .............> x = 11

Part b .............> k = 57.2

Part c .............> y = 9.2

<u><em>Explanation:</em></u>

The three problems deal with inverse variation between two variables

An inverse variation relation between two variables means that when one of the variables increases, the other will decrease (and vice versa)

<u>Mathematically, an inverse variation relation is represented as follows:</u>

$y = \frac{k}{x}$

where x and y are the two variables and k is the constant of variation

<u><em>Now, let's check the givens:</em></u>

<u>Part a:</u>

We are given that y = 3 and k = 33

<u>Substitute in the original relation and solve for x as follows:</u>

$y = \frac{k}{x}\\ \\3 = \frac{33}{x}\\ \\x=\frac{33}{3}=11$

<u>Part b:</u>

We are given that y = 11 and x = 5.2

<u>Substitute in the original relation and solve for k as follows:</u>

$y=\frac{k}{x}\\ \\11=\frac{k}{5.2}\\ \\k=11*5.2=57.2$

<u>Part c:</u>

We are given that x=7.8 and k=72

<u>Substitute in the original relation and solve for y as follows:</u>

$y=\frac{k}{x}=\frac{72}{7.8}=9.2$ to the nearest tenth

Hope this helps :)

6 0

3 years ago

the lenghty of a rectangal is 8 feet more than its width. the pirimete of the rectangle is 72 feet. Find the width

morpeh [17]

Width of the rectangle is 14 ft

Step-by-step explanation:

Step 1: Let the width of the rectangle be x. Then length = 8 + x. Perimeter = 72 ft.

Perimeter = 2(length + width)

72 = 2 (8 + x + x)

72 = 16 + 4x

4x = 56

x = 56/4 = 14

∴ Width of the rectangle is 14 ft

3 0

3 years ago

Help w algebra please

hjlf

The correct answer is C

6 0

3 years ago