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3 years ago

Please help me with this slope question!

Mathematics

2 answers:

Juliette [100K]3 years ago

8 0

The correct answer is y=2/3x

To find the slope you use rise over run so you take one dot (2,-5) and go up 2 units so that it would get one the same line as the other dot (5,-3) and then go to the right 3 units to get you to the exact spot that the dot is

galben [10]3 years ago

8 0

Answer:

2/3

Step-by-step explanation:

the points are (2,-5) and (5,-3)

the slope is change in y over change in x

so the slope can be written as:

(-3-(-5))/(5-2)

thats 2/3

and thats the slope

:D hope this helps u

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(i) Given that

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v[r_] := 0.30 - 3.33r^2

and additionally define the volume function with

V[R_] := Integrate[2 Pi v[r] r, {r, 0, R}]

Then get the desired volume by running V[0.30].

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$\displaystyle \int_0^R 2\pi K(R^2-r^2)r \, dr$

Compute the integral:

$V(R) = \displaystyle \int_0^R 2\pi K(R^2-r^2)r \, dr$

$V(R) = \displaystyle 2\pi K \int_0^R (R^2r-r^3) \, dr$

$V(R) = \displaystyle 2\pi K \left(\frac12 R^2r^2 - \frac14 r^4\right)\bigg_0^R$

$V(R) = \displaystyle 2\pi K \left(\frac{R^4}2- \frac{R^4}4\right)$

$V(R) = \displaystyle \boxed{\frac{\pi KR^4}2}$

In M, redefine the velocity function as

v[r_] := k*(R^2 - r^2)

(you can't use capital K because it's reserved for a built-in function)

Then run

Integrate[2 Pi v[r] r, {r, 0, R}]

This may take a little longer to compute than expected because M tries to generate a result to cover all cases (it doesn't automatically know that R is a real number, for instance). You can make it run faster by including the Assumptions option, as with

Integrate[2 Pi v[r] r, {r, 0, R}, Assumptions -> R > 0]

which ensures that R is positive, and moreover a real number.

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