Question

Offering 50 points!!

Answer 1

Answer: Here is the answers to your previous question that was deleted. As for this question, I believe the answer is the first one.

1)
`200\pi`
2)
`3√(3)`
3)
`-1`
4) 9
5) 0.151

Explanation:

#1

`=(dV)/(dt) =(4)/(3) \pi .3r^3(dr)/(dt)`

`=(dV)/(dt) =4\pi r^2(dr)/(dt)`

`=(dV)/(dt) =4\pi r^2*2...[(dr)/(dt) =2]`

`=(dV)/(dt) = 8\pi r^2~~~ in^3/min`

`r=5`

`=(dV)/(dt) =8\pi (5)^2`

`=200\pi ~~~in^3/min`

#2

`(d)/(dt) (x^2+y^2)=(d)/(dt) (16)`

`2x((dx)/(dt) )+2y((dy)/(dt))=0`

`2x((dx)/(dt) )+-2y((dy)/(dt))=0`
Divide both sides by 2.

`x((dx)/(dt))=-y((dy)/(dt))`

`(dx)/(dt)=(-y((dy)/(dt)) )/(x)`

As the particle is passing through the point
`(2,2√(3))` , So, the point will satisfy the equation. Substituting the value of
`x=2` ,
`y=2√(3)` and
`(dy)/(dt) -3`,

we get ,

`(dx)/(dt) =(((-2√(3))*(-3)))/(2)`

`=(6√(3) )/(2)`

`=3√(3)`

#3

Go on to Desmos and put in
`x+3y=6` and look at x=0 and x=1 and use those two as your points.

`(y_2-y_1)/(x_2-x_1)`

`=(1-3)/(2-0)`

`=(-2)/(2)`

`=-1`

#4

`y=3x^2-4x-6`

`(dy)/(dt) =5x-6`

`(dy)/(dt)=5(3)-6`

`(dy)/(dt) =9`

#5

`V=s^3`

`(dV)/(dt) =3*s^2((ds)/(dt))`

`-2=3*s^2((ds)/(dt) )~~~~~~~~~~~~~Equation 1`

`(ds)/(dt) =(1.33)/(s^2)`

`V=s^3`

`(27)^0^.^3^3=(s^3)^0^.^3^3`

`s` ≈
`2.97 ~~~~~~~~~~~~~~~~~~~~~~Equation2`

Substitute equation2 to equation1, we get

`(ds)/(dt) =(1.33)/((2.97)^2) =0.151`

I hope this helps you! Cheers! ^^