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3 years ago

What is the y intercept of the given graph?

Mathematics

1 answer:

Genrish500 [490]3 years ago

6 0

Answer:

The answer would be -2.

Step-by-step explanation:You have to look at where the line is hitting.It has to be perfectly on the corners of the square. The bottom of the y-intercept is always negative and the top is always positive but for the x-intercept the right side is positive and the left side is negative. I hope that helped you!!

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3 years ago

The cost to rent a jet ski is $80 per hour. in addition there also includes fee for a lesson on how to use the jet ski. Steve re

melomori [17]

Answer:

$y=80x+25$

Step-by-step explanation:

Let

x ----> the number of hours

y ---> the total cost of renting a jet ski in dollars

we know that

The linear equation in slope intercept form is equal to

$y=mx+b$

where

m is the slope or unit rate of the linear equation

b is the y-intercept or initial value

In this problem we have

The slope is equal to

$m=\$80\ per\ hour$

we have the ordered pair (3,265)

substitute in the linear equation

$265=80(3)+b$

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$b=265-240$

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substitute

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3 years ago

G find the area of the surface over the given region. use a computer algebra system to verify your results. the sphere r(u,v) =

Svetach [21]

Presumably you should be doing this using calculus methods, namely computing the surface integral along $\mathbf r(u,v)$ .

But since $\mathbf r(u,v)$ describes a sphere, we can simply recall that the surface area of a sphere of radius $a$ is $4\pi a^2$ .

In calculus terms, we would first find an expression for the surface element, which is given by

$\displaystyle\iint_S\mathrm dS=\iint_S\left\|\frac{\partial\mathbf r}{\partial u}\times\frac{\partial\mathbf r}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

$\dfrac{\partial\mathbf r}{\partial u}=a\cos u\cos v\,\mathbf i+a\cos u\sin v\,\mathbf j-a\sin u\,\mathbf k$
$\dfrac{\partial\mathbf r}{\partial v}=-a\sin u\sin v\,\mathbf i+a\sin u\cos v\,\mathbf j$
$\implies\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}=a^2\sin^2u\cos v\,\mathbf i+a^2\sin^2u\sin v\,\mathbf j+a^2\sin u\cos u\,\mathbf k$
$\implies\left\|\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}\right\|=a^2\sin u$

So the area of the surface is

$\displaystyle\iint_S\mathrm dS=\int_{u=0}^{u=\pi}\int_{v=0}^{v=2\pi}a^2\sin u\,\mathrm dv\,\mathrm du=2\pi a^2\int_{u=0}^{u=\pi}\sin u$
$=-2\pi a^2(\cos\pi-\cos 0)$
$=-2\pi a^2(-1-1)$
$=4\pi a^2$

as expected.

6 0

3 years ago