CPhill

C) The bases of a cylinder are circles.

SA  = 

2pi *r * h  +  2 pi * r^2  =

2pi * r  * [ h + r ]  =

2 * 3.14 *  9  * [ 12 + 9 ]  =  1186.9 units^2

How  about

Output  =   4n  + 7   ????

Here's how this is obtained

Slope  =     [ 15 - 11] /  [ 2 - 1 ]   =   4/ 1 =  4

So...using the slope and any  (input, ouput)  point - I'll use (1, 11) -  we have

Output  =  4 ( n - 1) + 11

Output  = 4n - 4 + 11

Output  = 4n + 7

Output  =  4n

If this is the function........y = (x - 1)^2  + 3

None of the answers are correct

This is the graph of x^2   shifted to the right 1 unit and up 3 units...

Here's the proof :

https://www.desmos.com/calculator/frdqngrnhj

In the form

y = a ( x + h)  + ( - k )   the vertex is  ( -h , -k)

So...we have

y = 3(x + 2)^2  + ( - 3 )

So....the vertex  is ( -2, -3)

Only one answer fits.....the second one

y = 2(x - 3)^2 +   (-  4 )

Because the "2" out front is positive, this parabola turns upward.......the vertex  is  (3, -4)

So....this is the minimum point on the parabola.....and y =  -4  is the minimum

The domain is all real numbers  [  we can put any x value into the function ]

And the range is  [ -4, infinity]   ⇒ all real numbers ≥ 4

The first answer is correct

This graph will help you see this :

https://www.desmos.com/calculator/cmyimyjq0h

In the given ratio,  A = 2B   and C  = 4B

So  

[ 3 (2B) + 2B ]  / [ 4(4B) - 2B ] =

[ 8B ] / [ 14 B ]  =

8 / 14  =

4 / 7

If we let x = 0, then y =  2(0)^2  = 0...so  (0,0)  will be on the graph

This eliminates the middle two graphs

If we let x  = 1, then y =  2(1)^2  =  2

So...( 1 , 2)   is on the graph

So...the first graph is correct.....

1. -2 ( a + 6)  =  -2a  - 2*6  =   -2a  - 12

2.  Whole Numbers, Rational Numbers, Integers 

3. "6 Times the difference of B and p?"  =   6 [ b -  p ]

4. 5(14-2)^2/2  =  5 (12)^2 / 2  =  5 * 144/2  =  5 * 72 =  360

5.  u + xy, if u= 18 and x= 10, AND y=8

    18  + (10) (8)  =

    18  +  80  =

        98

Here's the graph, NSS :

https://www.desmos.com/calculator/fgs3orbcai

The max will be found at a corner point...there's only one at (1.5, 1.5)