CPhill

We could do some math here....but inspection may be faster

Notice that the corresponding sides of the larger triangle are 3  times that of the smaller triangle

So....the missing side in the larger triangle must be 3 times 7  =   21  = x

x^2 + y^2 - 4x + 2y - 11  = 0

Rearrange as

x^2- 4x + y^2 + 2y  =  11

We need to complete the square on x  an y 

I'm assuming that you know how to do this...if, not...here's a good resource :

http://www.purplemath.com/modules/sqrcircle.htm

x^2 - 4x + 4 + y^2 + 2y + 1  =  11  + 4  + 1       factor and simplify

(x - 2)^2  +  (y + 1)^2   =  16

In the form    ( x - h)^2  + (y - k)^2  = r^2

The center  is  (h, k)   and the radius  = r  ......so....we can write

( x - 2)^2  +  ( y  - (-1) )^2  =   4^2

The center is   (2, - 1)   and the radius  is  4

2.   The center is  (-1,1)

a) Note that the point  (-1,4)  is also on the graph...and the distance between this point and the center is jthe radius....this is given by the difference of the y coordinates   ( 4 - 1)  =  3

b) 

Graphing form  

 (x + 1)^2  + (y - 1)^2  =   9

Gerenral form...expand the graphing form

x^2 + 2x + 1 + y^2 - 2y + 1   = 9

x^2 + y^2 + 2x - 2y + 2  =  9           subtract 9 from both sides

x^2 + y^2 + 2x - 2y  - 7   =  0

3.

a)   

 In the form  ( x - h)^2  + (y - k)^2  = r^2    .... we can write this as

(x - 0)^2 + ( x -  (-5) )^2   = 9

The center  is  ( h , k)  =  ( 0 , -5)    ....so Jenna is incorrect

b)   9  =  r^2  ...so   the radius is the square root of this     = 3 units

So....the diameter is twice this  =  6 units

Raul is incorrect, as well  !!!

We need to find the radius...

The surface area is given by :      4 pi * r^2

So

16 pi  =  4 pi * r^2     divide both sides by pi

16  = 4r^2     divide both sides by 4

4  = r^2       take the positive root

2  ft  = r

So...the volume of a sphere is

(4/3) pi ( radius)^3  =

(4/3) pi (2) ^3

(4/3) pi * 8

(32/3) pi   ft^3   ≈   10.66 pi  ft^3

None of the given answers are correct  !!!

1.  Once we fold the sides up....the dimensions of the  base will be

(8 - 2x) (10 - 2x)

And the height will be x

So....the volume, V, is   area of the base * height  =

a)  V = (8 - 2x) (10 - 2x) x

b)  Here's a graph   of the function :

https://www.desmos.com/calculator/ubq8tavuol

The value of x that gives the max volume is when we look at the top of the curve between x  = 0  and x = 4....this value is :   x  ≈ 1.5  in

2. 

a) The volume of the aquarium is  s^3  =  3^3  =   27 ft^3

The weight will be    27 * 62  = 1674 lbs.     the table will NOT support that !!!

b)  The density of water is the same, no matter the volume

Sorry, cerenetie......no clue  !!!

2.What is the area of the region defined by the equation $x^2+y^2 - 7 = 4y-14x+3$?

Rewrite  as

x^2 + 14x +y^2 - 4y = 10

This is a circle....we need to complete the square on x and y  [ I'm assuming that you are familiar with this?? ]

x^2 + 14x + 49  + y^2 - 4y + 4  =  10 + 49 + 4

The right side simplifies to  63  = ( radius of the circle)^2

So... the area  is    pi * r^2   =  pi *63  ≈  197.9  units^2

I have 2 question

1.The graph of the quadratic $y = ax^2 + bx + c$ has the following properties: (1) The maximum value of $y = ax^2 + bx + c$ is 5, which occurs at $x = 3$. (2) The graph passes through the point $(0,-13)$. If the graph passes through the point $(4,m)$, then what is the value of $m$?

2.What is the area of the region defined by the equation $x^2+y^2 - 7 = 4y-14x+3$?

1. The vertex  (3, 5)  and the point  (0, - 13)  are on the graph ....and  (3,5)  is tthe vertex

The x coordinate of the vertex is given by :

-b / [2a]  =  3

-b = 6a

b = - 6a

And  since  (0, - 13)  is on the graph, then  c  =  -13

And using the vertex, we can find "a"  thusly

5  =  a(3)^2  - 6a(3)   - 13

18  = 9a- 18a

18  =  -9a

-2  = a       ⇒   b  = -6(-2)  =  12

So...our function is

y   =  -2x^2 + 12x - 13

And when x  =  4, m  =

-2(4)^2  + 12(4)  - 13

-32 + 48  - 13

3

So....the point (4,3)  is on the graph

Here's a graph : https://www.desmos.com/calculator/jhmml2xcki