Question 14 of 20 :

Ostrovityanka [42]

Sent a picture of the solution to the problem (s). I usually graph at least 5 points. Usually one is 0 and I go up from zero and down from zero.

7 0

3 years ago

15) Solve for X X\42=6 A) 7 B) 252 C) 260 D) 332

Vikentia [17]

Answer:

b)252 is the answer

Step-by-step explanation:

x\42=6

x=6*42

x=252

5 0

2 years ago

According to the US department of labor, total employment is expected to increase from 146 million in 2000 to 168 million in 201

saveliy_v [14]

Answer:

The percent increase in the employment from the year $2000$ to year $2010$ is $15.07\%$ .

Step-by-step explanation:

Given:

Employment in the year $2000$ $=146$ million.

Employment in the year $2010$ $=168$ million.

To find: The percent increase in the employment.

Solution: We have,

Employment in the year $2000$ $=146$ million.

Employment in the year $2010$ $=168$ million.

Increase in employment $=168-146=22$ million.

Percent increase $=\frac{\textrm{Increase in employment}}{\textrm{Original employment}}\times100\%$

Percent increase $=\frac{22}{146}\times100\%=15.07\%$

Hence, the percent increase in the employment from the year $2000$ to year $2010$ is $15.07\%$ .

3 0

3 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

What is the slope of a line perpendicular to the

Dahasolnce [82]

Answer: a. -3/5

Step-by-step explanation:

7 0

3 years ago

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