Me ! a high-altitude spherical weather balloon expands as it rises, due to the drop in atmospheric pressure. suppose that the radius r increases at the rate of 0.16 inches per second and that r = 38 inches at time t = 0. determine the equation that models the volume v of the balloon at time t, and find the volume when t = 280 seconds.

Find r as a function of time t.

r starts at 38 inches and steadily increases at 0.16 inch per second. So in t seconds r would have increased by 0.16t when added to the initial 38 inches will give:

r(t) = 38 + 0.16t

Volume of a sphere = 4/3(pi)r^3. Since radius changes with time, volume will change with time too.

V(t) = 4/3(pi)r^3 = 4/3(pi)(38 + 0.16t)^3

Put t=280 and solve to get V

V=2377823.53

step-by-step explanation:

first, eliminate the parenthesis:

3

x

+

2

−

4

x

second, group like terms:

3

x

−

4

x

+

2

third, combine like terms:

(

3

−

4

)

x

+

2

→

−

1

x

+

2

→

−

x

+

2

y = 3x+4

step-by-step explanation:

to find the slope

slope = (y2-y1)/ (x2-x1)

    = (-8-4)/ (-4-0)

    = -12/-4

  = 3

using the slope intercept form of a line

y= mx+b

we know the slope is 3

we know the y intercept is 4 ( the point where it crosses the y axis or x=0)

y = 3x+4