Assignment 14

course Mth 158

?????????L?}?assignment #014014. `query 14

College Algebra

07-01-2008

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20:18:38

**** query 1.6.12 (was 1.6.6). Explain how you found the real solutions of the equation | 1 - 2 z | + 6 = 0.

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RESPONSE -->

| 1-2z| +6 =0

1-2z +6 -6 = 9 -6

1-2z= 3

1 -1 -2z=4 Divide each side by -2

z=2

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20:26:15

** Starting with

| 1-2z| +6 = 9 we add -6 to both sides to get

| 1 - 2z| = 3. We then use the fact that | a | = b means that a = b or a = -b:

1-2z=3 or 1-2z= -3 Solving both of these equations:

-2z = 2 or -2z = -4 we get

z= -1 or z = 2 We express our solution set as

{-2/3,2} **

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RESPONSE -->

I needed to produce two equations instead of going through and solving the one. If the answers to the equations is -1 and 2. Why is the solution set (-2/3, 2)?

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20:36:42

**** query 1.6.30 (was 1.6.24). Explain how you found the real solutions of the equation | x^2 +3x - 2 | = 2

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RESPONSE -->

x^2 +3x -2 = 2 or x^2 + 3x -2 = -2

x^2 +3x -4 = 0 X^2 + 3x = 0

(x-1) (x+4) or x(x+3)

x-1 = 0 or x+4 = 0 or x=0 or x+3=0

x= 1 or x=-4 or x=0 or x=-3

(0, 1, -3, -4)

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20:37:37

** My note here might be incorrect.

If the equation is | x^2 +3x -2 | = 2 then we have

x^2 + 3x - 2 = 2 or x^2 + 3x - 2 = -2.

In the first case we get x^2 + 3x - 4 = 0, which factors into (x-1)(x+4) = 0 with solutions x = 1 and x = -4.

In the second case we have x^2 + 3x = 0, which factors into x(x+3) = 0, with solutions x = 0 and x = -3. **

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RESPONSE -->

ok that is what I had

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20:42:32

**** query 1.6.40 \ 36 (was 1.6.30). Explain how you found the real solutions of the inequality | x + 4 | + 3 < 5.

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RESPONSE -->

|x+4| +3 < 5

|x+4| < 2

-2 < x+4 < 2

-6 < x < -2

-6 > x >-2

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20:42:48

STUDENT SOLUTION: | x+4| +3 < 5

| x+4 | < 2

-2 < x+4 < 2

-6 < x < -2

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RESPONSE -->

ok

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20:48:37

**** query 1.6.52 \ 48 (was 1.6.42). Explain how you found the real solutions of the inequality | -x - 2 | >= 1.

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RESPONSE -->

| -x-2| > = 1

1 <-x -2 < = 1

3<-x < 3

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20:49:11

**Correct solution:

| -x -2 | >= 1 Since | a | > b means a > b or a < -b (note the word 'or') we have

-x-2 >= 1 or -x -2 <= -1. These inequalities are easily solved to get

-x >= 3 or -x <= 1 or

x <= -3 or x >= -1.

So our solution is

{-infinity, -3} U {-1, infinity}. **

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RESPONSE -->

ok

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W????????????

assignment #015

015. `query 15

College Algebra

07-01-2008

?????T?????€?assignment #015

015. `query 15

College Algebra

07-01-2008

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21:11:15

**** query 1.7.20 (was 1.2.30). Explain how you set up and solved an equation for the problem. Include your equation and the reasoning you used to develop the equation. Problem (note that this statement is for instructor reference; the full statement was in your text) million to lend at 19% or 16%, max lent at 16% to average 18%.

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RESPONSE -->

I need help, I could not get my equation set up for this one.

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If you let x = the amount lent at 19% then how much is lent at 16%

What is an expression for the interest on amount x at 19%?

What is an expression for the interest on the amount lent at 16%?

What must be the total interest?

You should get an equation solvable for x. What is your equation and what is your solution? What therefore is your solution to the problem?