Complete the function for this graph.

Find the product of 3 and the linear equation, and solve for both equations for y.

krok68 [10]

Answer:

Part 1) The product of 3 and 2y-4x=8 is $6y-12x=24$

Part 2) The original equation in slope intercept form is $y=2x+4$

Part 3) The new equation in slope intercept form is $y=2x+4$

Step-by-step explanation:

we have

$2y-4x=8$

step 1

Find out the product of 3 and 2y-4x=8

Multiply both sides by 3

$3(2y-4x)=3(8)$

Apply distributive property both sides

$6y-12x=24$

step 2

Find out the original equation in slope intercept form

The equation of the line in slope intercept form is

$y=mx+b$

we have

$2y-4x=8$

Solve for y

That means ----> Isolate the variable y

Adds 4x both sides

$2y=4x+8$

Divide by 2 both sides

$y=(4x+8)/2$

Simplify

$y=2x+4$

step 3

Find out the new equation in slope intercept form

The equation of the line in slope intercept form is

$y=mx+b$

we have

$6y-12x=24$

Solve for y

That means ----> Isolate the variable y

Adds 12x both sides

$6y=12x+24$

Divide by 6 both sides

$y=(12x+24)/6$

Simplify

$y=2x+4$

8 0

3 years ago

A sphere of radius r is cut by a plane h units above the equator, where

Anika [276]

Consider the top half of a sphere centered at the origin with radius $r$ , which can be described by the equation

$z=\sqrt{r^2-x^2-y^2}$

and consider a plane

$z=h$

with $0$ . Call the region between the two surfaces $R$ . The volume of $R$ is given by the triple integral

$\displaystyle\iiint_R\mathrm dV=\int_{-\sqrt{r^2-h^2}}^{\sqrt{r^2-h^2}}\int_{-\sqrt{r^2-h^2-x^2}}^{\sqrt{r^2-h^2-x^2}}\int_h^{\sqrt{r^2-x^2-y^2}}\mathrm dz\,\mathrm dy\,\mathrm dx$

Converting to polar coordinates will help make this computation easier. Set

$\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\var\phi\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi$

Now, the volume can be computed with the integral

$\displaystyle\iiint_R\mathrm dV=\int_0^{2\pi}\int_0^{\arctan\frac{\sqrt{r^2-h^2}}h}\int_{h\sec\varphi}^r\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta$

You should get

$\dfrac{2\pi}3\left(r^3\arctan\dfrac{\sqrt{r^2-h^2}}h-\dfrac{h^3}2\left(\dfrac{r\sqrt{r^2-h^2}}{h^2}+\ln\dfrac{r+\sqrt{r^2-h^2}}h\right)\right)$

5 0

3 years ago

Help me please summer homework

vladimir1956 [14]

86. Is a negative slope because it slants down from left to right.
87. Is positive slope because it slants up from left to right.
88. Is an undefined slope because it is vertical.
89. Is has a slope of zero because it is horizontal.

8 0

1 year ago

Justin wants to use 376 ft of fencing to fence off the greatest possible rectangular area for a garden. what dimensions should h

Rina8888 [55]

Let the length be x and the width be y
the perimeter of the rectangle will be:
2x+2y=376
thus
y=188-x
thus the area will be:
A(x)=x(188-x)
A(x)=188x-x^2

For maximum Area A'(x)=0
from Area we shall have
A'(x)=188-2x=0
solving for x we get:
x=94
thus the width will be:
188-94=94
thus for maximum area the length=94 ft and width=94 ft
Area=94*94=8836 ft^2
thus the answer is:
<span>c. 94 x 94; 8836ft </span>

3 0

3 years ago

Read 2 more answers

Solve the system of equations: y²-x²=16, x²-y²=16 (explain and show work) Will name most brainy

mr Goodwill [35]

<h3>Answer: No solutions (system is inconsistent)</h3>

=====================================================

Explanation:

Rearrange the first equation into $-x^2+y^2 = 16$

So we have this equivalent system

$\begin{cases}-x^2+y^2 = 16\\\\x^2-y^2=16\end{cases}$

If you add the terms straight down, then you'll find that the x^2 and y^2 terms add to 0. The right hand side terms add to 16+16 = 32

We are left with the equation 0 = 32, which is a false equation or contradiction. Therefore, there are no solutions. We say the system is inconsistent. The two graphs do not intersect at all as shown in the diagram below. We have two hyperbolas in which the branches extend off to infinity to slowly approach the asymptotes. They never actually get to the asymptotes, but only get closer.

7 0

3 years ago