Ahigh-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 0.02 inches per second, and that r = 36 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 = 360 seconds.

v(t) = 4π(0.02t)2; 651.44 in3

v(t) = 4π(36 + 0.02t)2; 1,694,397.14 in3

v(t) = four pi times the product of zero point zero two and t to the third power divided by three.; 4,690.37 in3

v(t) = four pi times the quantity of thirty six plus zero point zero two t to the third power divided by three.; 337,706.83 in3

V(t)=

337 706, 83

Step-by-step explanation:

The volume of a sphere is

So like the initial radius is 36 and is increasing of 0.02 every second, we define the radius:

r= 36+0.02t

So, we replace in the formula of the volume and we obtain

and if we replace t= 360 seconds in the equation of volume, we get

answer: 75°

step-by-step explanation:

∠esk is the supplement of ∠edk, so is 100°. ∠esd = 1/2·arc ed, so is 25°.

since ∠esk = ∠esd + ∠dsk

you have 100° = 25° + ∠dsk . .

fill in known values 75° = ∠dsk . . . . .

subtract 25°

step-by-step explanation:

step by step resolutions

1) n+n+1+n+2=51

3n+3=51

3n=48

n=16

the three numbers are 16, 17 and 18.

2) x+15=4x

3x=15

x=5

radha is now 5 years old

3) the two digits are x and y.

y=3x

(10x+y)+(10y+x)=88

y=3x

11x+11y=88

y=3x

x+y=8

x+3x=8

x=2

the original number can be either 28 or 82.

4) n+(n+2)+(n+4)=48

3n+6=48

3n=42

n=14

the numbers are 14, 16 and 18.

5) y=x+3

10y+x+10x+y=143

y=x+3

11x+11y=143

y=x+3

x+y=13

x+x+3=13

2x=10

x=5

the original number can be 57 or 75.

hope this !