CPhill

\( h(x) = \sqrt{25-x^2}+\sqrt{-(x-2)}+\sqrt{x}\)

h(x)  =  sqrt (25 - x^2)  + sqrt (2 - x) +  sqrt (x)

The first term  has a domain  of   [ -5 , 5 ]

The second term has a domain of   ( -inf , 2]

The third term has a domain of  [ 0 , inf)

The most restrictive interval   is   [ 0 ,2 ]

So....the width of the domain interval  =  2

By the Triangle inequality

7 + 12 >  x              and      7 + x  > 12

19 > x                                 x > 12 - 7

x < 19                                 x > 5

So the possible values of  x  are         5  < x < 19

Derivative  of   - 3.11* t^2  =   2(-3.11) * t^(2-1)  =  - 6.22 * t  

Derivative of  9.33t^1  =   1*9.33* t^(1 - 1)  = 9.33t^0  = 9.33

General rule

Derivative of  a *t^n  =   n*a * t ^(n -1)

B + A   =100

B = .25A

So

.25A  + A  = 100

1.25A =  100

A = 100 /1.25 =   80

B = .25A = .25(80) = 20

Blake raised  80  - 20   = $60  less

Call S  the original amt

Amount left at the end of Oct =  .7S

Amount left at the end of Nov =   .7S / 5 = .14S

After buying two calculators  and giving (3/8)  of the money left to his parents he had 

(.14S - 2(15.80)) - (3/8)[ .14S - 2(15.80) ]  =

.14S  - 31.60  - .0525S + 11.85 =

.0875S - 19.75     left

In Dec  he had    .40 [ .0875 S  - 19.75 ]  left

And this equals  6219.85

So

.40 [ .0875S - 19.75] =  6219.85

.0875S  - 19.75 =  6219.85  /.40

.0875S  -19.75 = 15,549.63

.0875S  =  15,549.63 + 19.75

.0875S  = 15,569.38

S  =   15569.38/.0875   =  $ 177,935.77

He gave his parents  (3/8)[ .14 * 177935.77 - 31.60 ]  = $9329.78

Cal N the total number baked

Monday she baked  68

Tuesday she baked (3/8)(N - 68)

Wed she baked  65

So

68  + (3/8)(N -68) + 65 = N

133 + (3/8)N  -  (3/8)(68)  = N            rearrange as

133  - 25.5   =  (5/8)N         

107.5 = (5/8)N

N =  107.5 ( 8/5)   =     172

Assuming repeats :  Number of possible  "words" with no restrictions =   (26)(26)  = 26^2 =  676

Again, assuming repeats......Number of possible words with only  consonants =  (21)(21) =  441

Number of words with at least one  vowel  =   676  - 441   =  235

z =  3 + 4i  + 5 + 6i  =   8 + 10i

z^2  =  ( 8 + 10i)^2  =   64 + 160i  + 100i^2  =    64  + 160i  + 100 (-1)   = 64  + 160i - 100  =

-40  + 160i

If 3 + 2i  is a root  then so is 3 - 2i

By Vieta

ax^2 + bx + c  = 0

Sum of the roots =  -b/a           and a  =1

So   (3 + 2i)  + (3 -2i)  = 6/1   = -b/1   → b = -6

Product of the roots =   c/a

So (3+2i)(3-2i)  =  9 - 4i^2  =  9 -4(-1)  = 9 + 4 = 13  =    13/1  →   13  =  c  

1) The function is

h(t)  = -16t^2  + vt  + s   =

h(t)  = -16t^2 + 63t  

2)  Here's the graph  (in x instead of t)  : https://www.desmos.com/calculator/96nhz6lqgq

3) Note that with the given function, the ball never rises above ≈  62 ft

4) If the punter's initial velocity is 68 ft/s we have the graph  of -16t^2 + 68t : https://www.desmos.com/calculator/2bn74gqnob

The ball will be > than the stadium height  between  1.75  and 2.125 sec