CPhill

Question 2

This says to  R1  and from this subtract  (1/2)R2    [and put the result in R2 ]

(1/2) R2  = (1/2)  [  -6   -3   12 ]  =  [ -3  -3/2     6]

Subtract this from R1....so we have

[ 4 - -3      3 - -3/2     0 - 6 ]   =  [ 7    9/2    - 6 ]  

So   the new matrix is

4    3       0

7   9/2    -6

First one says to take 2R2  and subtract R4     [ and I assume we put the result in R4 ]

So   2R2  =    2 [ 0   4 ] =   [ 0   8 ]

And from this subtract R4  =   [ 0 - -3 , 8 -  -4 ]   =  [ 3, 12]

So  the new matrix is

-1   2

0    4

6    3

3   12

Amount of coupon  = $15

The discount  off the price  =  30%......to see this.....the price is .7P  = 70% of the normal price, P...so....the discount must be =  .30   =  30%

1. There are two values of k for which the cubic polynomial 2x^3 - 9x^2 + 12x - k has a double root. What is the sum of those values?

Let  the  double root  be " a "   and the other root be " b"

And from Vieta, we have that

a + a + b =  2a + b   = 9/2    →   b  =  9/2 - 2a      (1)

a^2 + ab + ab  = a^2 + 2ab  =  6       (2)

a^2b =  k/2  →  2a^2b  = k      (3)

Sub (1)  into (2)  and we have that

a^2 + 2a ( 9/2 - 2a)  = 6

a^2 + 9a - 4a^2  = 6

-3a^2 + 9a  - 6  = 0

3a^2 - 9a + 6  = 0

a^2 - 3a + 2  = 0

(a - 2) (a - 1)  = 0

So    a = 2    or     a  =  1

If a = 1, then b =  (9/2) - 2a  =  9/2 -2(1)  =  5/2

And k  = 2(1)^2 (5/2)  =  5

And if a = 2, then b  = (9/2) -2(2)  = (9/2) - 4   =  1/2

And k  =  2(2)^2 ( 1/2)  =  4

So the sum  =  4 + 5  =  9  

Good job, tertre   !!!!

Thanks, CU   .....that  cubic "formula" is nasty......!!!!

We have more variables than equations....so.....we will have  infinite solutions

Subtracting the first equation from the second  we get that  3y + 5z  =  16  → 

 3y = [ 16 - 5z]  →     y =  [ 16 - 5z ] / 3 

And using the first equation

x + y + z =  8

x + [ 16 - 5z]/3 + z  = 8       multiply through by 3

3x + 16 -5z + 3z = 24

3x - 2z = 8

3x = 8 - 2z

x = [ 8-2z] /3

So....the solutions for  {x, y , z }  =   {  [8-2z]/3 , [16-5z]/3 , z }

For example...if z = 0,  then x = 8/3  and y = 16/3

So....check that these solutions work :

x + y + z  = 8  ???

8/3 + 16/3 + 0  = 8 ???

24/3 = 8  ???

True

And

x + 4y + 6z = 24  ???

(8/3) + 4 [16/3]  + 0  = 24??

8/3 + 64/3  = 24 ???

72/3  = 24   ???

True

So.....   one possible solution is   { x, y , z}  =  (8/3, 16/3 , 0 }

3. The fourth-degree polynomial P(x) satisfies P(1)=1, P(2)=2, P(3)=3, P(4)=4, and P(5)=125.What is P(6)?

A  fourth degree polynomial  has the form  ax^4 + bx^3 + cx^2 + dx + e

We have the following equations

a + b + c + d + e  =  1

16a + 8b + 4c + 2d + e  = 2

81a + 27b + 9c + 3d + e  = 3

256a + 64b + 16c + 4d + e  = 4

625a + 125b + 25c + 5d + e  = 125

This system is a little lengthy to solve....Wolfram Alpha gives the following solutions

a = 5  b = -50  c  = 175    d = -249   e  = 120

So.... P(6)  =   5(6)^4 - 50(6)^3 + 175(6)^2 - 249(6) + 120   =  606 

Nice, CU  !!!!

This uses the Exterior Angle Theorem

It says that

measure of DBA  = measure of BAC + measure of  ACB

124  = 95  + measure of ACB    subtract 95 from both sides

29°  =   measure of ACB