CPhill

When x  = -3,  then f(-3)  = - (-3)^2 + 4   =   - 9 + 4    =  -5

And from the table, when x  = - 3, then g(-3)  = 1

So  when x = -3, 

f(-3) < g(-3)

C. When the pile is changing at a rate of 1.5 feet per minute, how tall is the pile?

dh/ dt  =  4pi h^2

1.5 =  4 pi h^2

(1/5) / ( 4pi)  = h^2

√[ 1.5 / (4 pi) ]  = h ≈   .345 ft 

B. When the  pile is 5 feet high, how fast is the height of the  pile changing

dv / dh  = 4pi h^2

So......

dV / dt   =  dV/ dh * dh /dt

8 ft^3   = 4pi h^2 * dh/dt

8 ft^3  = 4pi (5ft)^2  * dh/dt

8 ft^3 = 100pi ft^2 * dh /dt

8 / [ 100pi ]  = dh/dt

.08/pi  ft /min  = dh/dt

No figure is shown.....

A.  

v(t)  = 4t(t - 7) (t - 1)     simplify

v(t)  = 4t  ( t^2 - 8t + 7)

v(t)  =  4t^3 - 32t^2 + 28t

The acceleration at any time  is the derivative of this  = v'(t)  = 12t^2 - 64t + 28

B.  Set the derivative to 0  and solve for t

12t^2 - 64t + 28  = 0        divide through by 4

3t^2 - 16t + 7     =  0         

The minimum acceleration on [0,3 ] occurs at  t  =    (16)/ (2 * 3)  =  16/6 =  8/3  sec

And the minimum acceleration is   3(8/3)^2  - 16(8/3) + 7   =  -43/3

C. See the graph here :  https://www.desmos.com/calculator/70cdiunnem

The max velocity on [0, 2 ] occurs at about .481 s   and is  ≈ 6.51

The power of the  polynomial  in the numerator is less than that in the denominator.....so.....the limit  of this series will be  0

We need to find  two  different primes  that add to  180  - 41  = 139°

One prime must  be even  =  2°

The other must be = 137°

The angles can only be    2, 41, 137

So  B =  137°  and C =  2°

You  can use  any letter you like for the  index   ( and function)

"i"  is as good as "n"

sec  and csc  are  co-functions

Specificallly

sec ( A)  =   csc (90 - A)

Note that

(A)  +  (90 -A)  = 90

So....this must mean that

(4x)  + (x - 20)   = 90

5x - 20  = 90   add 20 to  both  sides

5x = 110        divide both sides by 5

x = 22°