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What is 208 divided by 4

vodomira [7]

Answer:52

Step-by-step explanation:

5 0

3 years ago

Graph by using the slope and the y intercept.

Mashutka [201]

1. a) equation of the line :

$y = \dfrac{2}{5}x - 7$

y - intercept = -7

so, it will pass through point (0, -7)

and if we plug the value of x as 5, we get

$y = (\dfrac{2}{5} \times 5) - 7$

$y = 2 - 7$

$= - 5$

so, it will pass through point (5, -5) too

now, just plot the points (0 , -7) and (5 , -5) and join them.

2. b) equation of line is :

$y = - 3x + 5$

here, y - intercept = 5

so the line passes through point (0 , 5)

now, Plugging the value of x = 1 we get :

$y =( - 3 \times 1) + 5$

$y = - 3 + 5$

$y = 2$

so, the given line passes through point (1 , 2)

plotting the points, we can get our required line.

6 0

2 years ago

Suppose that \nabla f(x,y,z) = 2xyze^{x^2}\mathbf{i} + ze^{x^2}\mathbf{j} + ye^{x^2}\mathbf{k}. if f(0,0,0) = 2, find f(1,1,1).

lesya [120]

The simplest path from (0, 0, 0) to (1, 1, 1) is a straight line, denoted $C$ , which we can parameterize by the vector-valued function,

$\mathbf r(t)=(1-t)(\mathbf i+\mathbf j+\mathbf k)$

for $0\le t\le1$ , which has differential

$\mathrm d\mathbf r=-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt$

Then with $x(t)=y(t)=z(t)=1-t$ , we have

$\displaystyle\int_{\mathcal C}\nabla f(x,y,z)\cdot\mathrm d\mathbf r=\int_{t=0}^{t=1}\nabla f(x(t),y(t),z(t))\cdot\mathrm d\mathbf r$

$=\displaystyle\int_{t=0}^{t=1}\left(2(1-t)^3e^{(1-t)^2}\,\mathbf i+(1-t)e^{(1-t)^2}\,\mathbf j+(1-t)e^{(1-t)^2}\,\mathbf k\right)\cdot-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt$

$\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)(t^2-2t+2)\,\mathrm dt$

Complete the square in the quadratic term of the integrand: $t^2-2t+2=(t-1)^2+1=(1-t)^2+1$ , then in the integral we substitute $u=1-t$ :

$\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)((1-t)^2+1)\,\mathrm dt$

$\displaystyle=-2\int_{u=0}^{u=1}e^{u^2}u(u^2+1)\,\mathrm du$

Make another substitution of $v=u^2$ :

$\displaystyle=-\int_{v=0}^{v=1}e^v(v+1)\,\mathrm dv$

Integrate by parts, taking

$r=v+1\implies\mathrm dr=\mathrm dv$

$\mathrm ds=e^v\,\mathrm dv\implies s=e^v$

$\displaystyle=-e^v(v+1)\bigg|_{v=0}^{v=1}+\int_{v=0}^{v=1}e^v\,\mathrm dv$

$\displaystyle=-(2e-1)+(e-1)=-e$

So, we have by the fundamental theorem of calculus that

$\displaystyle\int_C\nabla f(x,y,z)\cdot\mathrm d\mathbf r=f(1,1,1)-f(0,0,0)$

$\implies-e=f(1,1,1)-2$

$\implies f(1,1,1)=2-e$

3 0

3 years ago

Examine the intersection of these two lines:

grin007 [14]

Answer:

You can't answer this question since there is no picture. Remember though the any 2 angles that make 180 degrees or resembles a line is 180 degrees.

7 0

3 years ago

A worker is being raised in a bucket lift at a constant speed of 3 ft/s. When the worker's arms are 10 ft off the ground, her co

sweet [91]

H=3t+10 and h=-16t+15t+6

They will never be at the same height at the same time

7 0

3 years ago

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