All categories

3 years ago

Assignment #7

Mathematics

1 answer:

max2010maxim [7]3 years ago

4 0

Answer:

Step-by-step explanation:

That is a factorial, which is calculated by multiplying 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 (which is a really BIG number!) 40,320 to be exact!

You might be interested in

3. Jason earns $400 per week plus a 3% commission on his total sales. If he sold $675 in total sales, how much did he earn this

dlinn [17]

Answer:

$420.25

Step-by-step explanation:

Total earnings of the week

$= 400 + 3 \% \: of \:675 \\ \\ = 400 + 0.03 \times 675 \\ \\ = 400 + 20.25 \\ \\ = \$ 420.25$

6 0

3 years ago

Can someone do #2?❤️

ohaa [14]

Answer:

Step-by-step explanation:

A proportional relationship is a straight line. Is must also go through the point (0,0)

3 0

3 years ago

Read 2 more answers

Surface integrals using an explicit description. Evaluate the surface integral \iint_{S}^{}f(x,y,z)dS using an explicit represen

Jobisdone [24]

Parameterize $S$ by the vector function

$\vec r(x,y)=x\,\vec\imath+y\,\vec\jmath+f(x,y)\,\vec k$

so that the normal vector to $S$ is given by

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=\left(\vec\imath+\dfrac{\partial f}{\partial x}\,\vec k\right)\times\left(\vec\jmath+\dfrac{\partial f}{\partial y}\,\vec k\right)=-\dfrac{\partial f}{\partial x}\vec\imath-\dfrac{\partial f}{\partial y}\vec\jmath+\vec k$

with magnitude

$\left\|\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}\right\|=\sqrt{\left(\dfrac{\partial f}{\partial x}\right)^2+\left(\dfrac{\partial f}{\partial y}\right)^2+1}$

In this case, the normal vector is

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=-\dfrac{\partial(8-x-2y)}{\partial x}\,\vec\imath-\dfrac{\partial(8-x-2y)}{\partial y}\,\vec\jmath+\vec k=\vec\imath+2\,\vec\jmath+\vec k$

with magnitude $\sqrt{1^2+2^2+1^2}=\sqrt6$ . The integral of $f(x,y,z)=e^z$ over $S$ is then

$\displaystyle\iint_Se^z\,\mathrm d\Sigma=\sqrt6\iint_Te^{8-x-2y}\,\mathrm dy\,\mathrm dx$

where $T$ is the region in the $x,y$ plane over which $S$ is defined. In this case, it's the triangle in the plane $z=0$ which we can capture with $0\le x\le8$ and $0\le y\le\frac{8-x}2$ , so that we have

$\displaystyle\sqrt6\iint_Te^{8-x-2y}\,\mathrm dx\,\mathrm dy=\sqrt6\int_0^8\int_0^{(8-x)/2}e^{8-x-2y}\,\mathrm dy\,\mathrm dx=\boxed{\sqrt{\frac32}(e^8-9)}$

5 0

3 years ago

Please show me how you find the measure of side b and what b measures

Dovator [93]

Answer:

b = 252 in

Step-by-step explanation:

cos theta = adjacent side over hypotenuse

cos 38 = b/ 320

Multiply each side by 320

320 cos 38= b/320 * 320

320 cos 38 = b

252.1634412 =b

To the nearest whole number

252 =b

5 0

3 years ago

Which of the following graphs does not represent a function of x?

hram777 [196]

I believe that it is graph 3 :) if it’s not right let me know.

8 0

3 years ago

Read 2 more answers