Solve. Check your solution.<br> -4+s= -68

Which function has a greater rate of change?

omeli [17]

Answer:

Function A has a rate of change of 5 and Function B has a rate of change of 4.5. Thus, Function A has a higher rate of change

Step-by-step explanation:

3 0

2 years ago

What is the closest estimate for 179 divided by 9

vazorg [7]

179 / 9 is about 20
exact = 19.888888

3 0

3 years ago

The office manager of a small office ordered 140 packs of printer paper based on average daily use, she knows that the paper wil

Mandarinka [93]

slope intercept form is y=mx+b

So slope intercept form is The office manager of a small office ordered 140 packs of printer paper based on average daily use, she knows that the paper will last about 80 days

(A) Lets make a table

X axis represents the Number of days paper used

y axis represents the packs of printer paper used

x y

days packs of printer paper used

0 0 (0 days , 0 packs used)

80 140 (in 80 days , 140 packs paper used)

The graph is attached below

(B) To find Equation of a line we use points (0,0) and (80,140)

$slope = \frac{y_2-y_1}{x_2-x_1} = \frac{140-0}{80-0} = \frac{7}{4}$

y intecept is (0,0)

so b= 4

Slope intercept form of a line i y=mx + b

m is the slope and b is the y intercept

So slope intercept form of line becomes

$y= \frac{7}{4} x$

Standard form is Ax + By =C

$y= \frac{7}{4} x$

Multiply both sides by 4

4y = 7x

Now subtract 7x on both sides

-7x + 4y =0 is the standard form

(c) To find packs of printer paper the manager expect to have after 30 days, Plug in 30 for x and find out y

$y= \frac{7}{4} x$

$y= \frac{7}{4}(30)= 52.5$

Total 140 packs of printer paper

52.5 packs of paper used

Packs of paper remaining after 30 days = 140- 52.5= 87.5

8 0

2 years ago

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

2 years ago

Drag and drop the expressions to write the equation of the line in slope intercept form. <br> Y= _+_

son4ous [18]

Answer:nonfat did vt. R th

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Solve. Check your solution. -4+s= -68