CPhill

The hard part here is selecting the correct graph

f(x) is the parabola  ....it is the same in both graphs

g(x)   =    sqrt (x) - 1

Note that when x = 0, y = -1

So.... (0, -1) is on the graph of g(x)

So.... the bottom graph is correct

The intersection  occurs at x = 4....so....this is the solution

Just look at the columns where  f(x) = g(x)

Notice these are      

3      3        and 

6     6

This means that  when x = 1,  f(x) = g(x)   

And when x = 11,  f(x) = g(x)

So.....the x values that make these functions equal  are  the solutions  =

1  and 11

I believe this is correct

P(x > 15 )  =  

0

P ( 2 < x < 5 )

59 miles /1 hour  x    5280 ft / 1 mile   x     1 hour /60 min    x   1m / 60 sec  =  

[ 59 * 5280 ] / [ 60 * 60 ]   =  86.5 ft/ sec  = 87 ft / sec

BTW....here's an easier formula to use to convert to ft /sec

88 ft /sec = 60 mph

So

59  x   88 /60  =   86.5 ft/sec

y= 8 cot(nΘ)

Using the chain rule

y '  =   -8csc^2(nΘ) * n    =   -8n csc^2(nΘ)

Blank 1   Given

Blank 2  Definition of midpoint

Blank 3   Segment addition postulate

Blank 4   Division Property of Equality

Blank 5   Reflexive Property

Blank 6   Consequence of steps 8, 9

Blank 7   Corresponding parts of similar triangles are proportional 

Blank 8   Substitution

Blank 9   Multiplication Property of Equality

Answered here :

https://web2.0calc.com/questions/in-the-figure-lt-4-is-an-exterior-angle-to-vaul_1

y = -3x - 1

y = x^2 - 3x - 4

Set the y's equal

x^2 - 3x - 4 =  - 3x - 1        rearrange as

x^2 - 3 = 0

This has two solutions for x   = ±√3

Here's the first part

v = u + at

Note that a = change in velocity / change in time

So...... change in velocity  / change in time =  (v  - u ) / t  

So we have that

v =   u + at

v = u +  [ ( v - u) / t ]  * t   

v = u + (v - u)

v = u + v - u

v =  v