CPhill

V  =  L  * W  * H ...so...

10x^3  + 47x^2  + 16x - 48  = [ 5x - 4] [ 2x + 3] * H

10x^3 + 47x^2  + 16x  - 48 =  [ 10x^2 - 8x + 15x -12 ] * H

10x^3 + 47x^2 + 16x  - 48 =  [ 10x^2 + 7x -12 ] * H

We can find the height by  dividing both sides by  [10x^2 + 7x -12] 

So

H  =   [10x^3 + 47x^2 + 16x  -48]  / [ 10x^2 + 7x  - 12 ]

Performing ploynomial division to find the height , we have

                                  x   +  4

10x^2  +  7x  - 12  [   10x^3  + 47x^2 + 16x  - 48 ]

                                   10x^3 +  7x^2  -12x

                                  ______________________

                                                  40x^2  + 28x  - 48

                                                  40x^2  + 28x  - 48

                                                  _________________

So...the height  =    [ x + 4 ]

6 (3x + 4)   

distribute the  6 across the terms in  parentheses

6 * 3x   +  6 * 4  

18x + 24

11)Ms. Forsythe gave the same algebra test to her three classes. The first class averaged 80%, the second class averaged 85%, and the third 89%. Together, the first two classes averaged 83%, and the second and third classes together averaged 87%. What was the average for all three classes combined? Express your answer to the nearest hundredth.

The average  =  total points / number of students

So....total points = average*number of students

Let the total number of students in  each class = x , y  and z

The total number of points the first class amassed  was   80x

The total number of points amased by the second class was  85y

The total number of points amassed by the third class = 89z

So...for the first two classes we have

[ 80x + 85y ] / [ x + y] = 83

80x + 85y = 83x + 83y  subtract  80x, 83y from both sides   

3x = 2y

x = 2/3y

And,,,for the second two classes we have

[ 85y + 89z ]  / [ y + z ]  =  87  

[  85y + 89z ]  = 87 [y + z]

85y + 89z = 87y + 87z      subtract  85y, 87z from both sides

2z = 2y 

y  = z

So...for the three classes.....

Total points by all three classes /  number of class members  = the average for all three classes

[ 80x + 85y  + 89z ] /  [ x + y + z ]   =

[sub for  x and z ] 

[80 (2/3)y  + 85y  + 89y]  /  [ (2/3)y  + y  + y ] 

y [ 80 (2/3)  + 85  + 89 ]  / [  y [( 2/3) + 1 + 1] ]       [cancel the y's ]

[ 80 (2/3) + 85 + 89]  [  8/3 ]

[ 160/3 + 85 + 89 ] /  [8/3] =

[ 160/3  + 255/3 + 267/3]  / [8/3]     multiply  top/bottom by 3

[160 + 255 + 267 ] / 8  = 

85.25  = average for all three classes

Thanks, guest for the correction  ...!!!!

If I understand the question correctly....20 possible winning rolls

If Mary and Jenns roll different numbers, the larger number cannot be a factor of the smaller

The only way that both win is if the numbers are the same, because a number is alway a factor of itself

67  is a prime...the only integers that divide it  are 1 and -1

So

2x - 23  = 1     and       2x   - 23  = -1

2x  = 24                       2x  = 22

x  =12                           x  = 11

So...the "x"  sum is   12 +  11    = 23

Oops...sorry I missed that, shreyas1......!!!

You are corect  6 (12 x 12)  =  6 (144)  =  864 in^2

x^2 + 27  = 0

x^2  = -27       take both roots

x =  sqrt (-27)    or  x   =  -sqrt (-27)

x =   i *sqrt (27)     or    x  =  -i * sqrt (27)

x =  i * sqrt  ( 9 * 3)   or  x   = -i * sqrt (9 * 3)

x = 3isqrt(3)   or x = -3isqrt(3)

x^2  + 4x  + 13    = 0      subtract 13 from both sides

x^2  +  4x   =  -13         take 1/2 of 4 = 2...square it =  4  add to both sides

x^2  + 4x  +  4 =  -13 + 4        factor the left...simplify the right

(x + 2)^2  = -9     take both roots

x + 2  = sqrt (-9)      or   x + 2  = -sqrt(-9)

x+ 2  = 3i               or     x + 2  = -3i       subtract 2 from both sides

x = -2 + 3i   or    x =  -2 - 3i

16x^2  + 9y^2

[ 4x  + 3iy ]  [ 4x  - 3iy ]