Ask question

Ask question

All categories

3 years ago

9

60,000 is one tenth of

1 answer:

Tatiana [17]3 years ago

4 0

600,000 u just have to multiply it by 10

You might be interested in

Use the power-reducing formulas to rewrite each of the expressions in terms of the first power of the cosine. Sin^4cos^4

Katen [24]

Answer:

1/128 * (3 - 4cos2x + cos4x)

Step-by-step explanation:

See attachment for steps

6 0

3 years ago

Can you please help me with this question

Bingel [31]

the answer is 27.54

7 0

4 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

Aramp is 17 feet long, rises 8 feet above the floor, and covers a horizontal distance of 15 feet, as shown in the figure.

bonufazy [111]

Answer:

The ratio of Tan B is

$\tan B= \dfrac{AC}{BC}=\dfrac{8}{15}$

OR

$\tan A= \dfrac{BC}{AC}=\dfrac{15}{8}$

Step-by-step explanation:

In Right Angle Triangle ABC

angle C = 90°

AB = Ramp = 17 feet

BC =Horizontal distance = 15 feet

AC = Height from floor = 8 feet

To Find:

Ratio of Tan B = ?

Solution:

In Right Angle Triangle ABC By Tangent Identity we have

$\tan B= \dfrac{\textrm{side opposite to angle B}}{\textrm{side adjacent to angle B}}$

Substituting the given values we get

$\tan B= \dfrac{AC}{BC}=\dfrac{8}{15}$

OR

$\tan A= \dfrac{BC}{AC}=\dfrac{15}{8}$

5 0

3 years ago

this picture represents 1-4 of the distance from maria house to the park which represents the whole distance?

saveliy_v [14]

There is no picture provided sorry.

4 0

3 years ago