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#176 Re: Help Me ! » Solving Quadratics by Factoring and Graphing » 2012-12-12 02:29:07

Actually it is the last one:
-4.9t^2 + 24.5t + 117.6 = 0

First time I did it, I got a postive 3 in the end and chose that as my answer. I must have made a mistake somewhere then?

#177 Re: Help Me ! » Solving Quadratics by Factoring and Graphing » 2012-12-11 22:35:41

My class shows me how to do #15, this website shows the exact same thing: since I can't post links, search on Google: Using Quadratic Formula to Find the Zeros of a Polynomial  and click the first link.

Here are my answers:
-3.30 (round up becomes -3)
1.22 (round up becomes 1)

#178 Re: Help Me ! » Multiplying and Dividing Polynomials » 2012-12-11 22:17:39

2(x – 1) + 3(x – 1) / (x – 1)

2(x – 1)/(x - 1) + 3(x – 1)/( x - 1)

For the first fraction:
so (x - 1) goes into (x - 1) one time making it 2(1). Now I want to stop right here and make sure that I'm doing this right.

#180 Re: Help Me ! » Solving Quadratics by Factoring and Graphing » 2012-12-11 19:23:12

Alright, let me try that my self and please do tell any mistakes I make along the way. I will also try #15.

x = 3 and x = -7

(x – 3) (x + 7)
x(x – 3)
7(x - 3)

(x^2 – 3x)
(7x - 21)

When adding together, you are taking away 7x from 3x which becomes -4x. This changes the sign and that means the correct equation is:

x^2 + 4x – 21

I see where I made my mistake, I didn’t change the sign!

#181 Re: Help Me ! » Solving Quadratics by Factoring and Graphing » 2012-12-11 00:09:21

anonimnystefy wrote:

Hi bobbym

Her 4) and 13) are also correct.

Hi anonimnystefy,
I'm a guy... so it would be a he, not a she lol

---

Hi bobyym,
Thanks for answering. I submited my work already though and it seems that all but #6 and #15 are correct (according to my teacher). Right now she wants me to show her the work I have done to get those answers. I believe those two may be wrong.

#182 Help Me ! » Solving Quadratics by Factoring and Graphing » 2012-12-10 17:18:10

demha
Replies: 16

I would very much appreciate the help on checking if these are correct smile

Solve the quadratic equations in questions 1 – 5 by factoring.
1.
Q. x2 – 49 = 0

A. x = -7, 7

2.
Q. 3x3 – 12x = 0

A. x = -2, 2

3.
Q. 12x2 + 14x + 12 = 18

A. x = -3/2, 1/3

4.
Q.  –x3 + 22x2 – 121x = 0

A. x = 11, 11

5.
Q. x2 – 4x = 5

A. x = -1, 5

6.
Q. Work backwards to write a quadratic equation that will have solutions of x = 3 and x = -7.

A. x2 – 4x - 21

7.
Q. Work backwards to write a quadratic equation that will have solutions of x = 12 and x = 2.

A. x2 – 14x + 24

8.
Q. Work backwards to write a quadratic equation that will have solutions of x = -1/2 and x = 4.

A. x2 – 7/2x -2
And if we want to get rid of the fraction:
2(x2 – 7/2x -2)
2x2 – 14/2x – 2
(2 goes into 14, 7 times)
Final Answer: 2x2 – 7 - 4

9.
Q. Write a quadratic equation that will have a solution of only x = 0. Note: this means there will be a double solution of x = 0.

A. x2 + 0x + 0

10.
Q. Write a quadratic equation that cannot be factored.

A. x2 + x + 3

11.
Q. The product of two consecutive positive integers is 72. Find the integers.

A.
n(n + 1) = 72
n2 + n = 72
n2 + n – 72 = 0
(n + 9)(n – 8) = 0
Solution becomes -9 and 8. Problem asks for two positive numbers. We reject -9. Since n = 8 and there is n+ 1, we do 8 + 1 which comes up to 9.
Answer: 8, 9

12.
Q. The product of two consecutive negative integers is 10506. Write a quadratic equation that you could solve to find the integers.

A.
n(n + 1) = 10506
n2 + n = 10506
n2 + n – 10506 = 0
(n + 103)(n – 102) = 0
Solution comes to an obvious -103 and 102. Problem asks for two negative numbers.
n + 1 = -103 + 1 = -102
Answers: -102, -103

13.
Q. The product of two consecutive odd integers is 63. Write a quadratic equation that you could solve to find the integers, then find the integers.

A.
n(n + 2) = 63
n2 + n = 63
n2 + n – 63 = 0
(n + 9)(n – 7)
Answer: -9, 7

14.
Q. A tennis ball is launched with an initial velocity of 24.5 m/s from the edge of a cliff that is 117.6 meters above the ground. Which quadratic equation could be used to correctly determine when the ball will hit the ground:
4.9t2 + 24.5t + 117.6 = 0

-4.9t2 - 24.5t + 117.6 = 0

-4.9t2 + 24.5t - 117.6 = 0

4.9t2 + 24.5t - 117.6 = 0

-4.9t2 + 24.5t + 117.6 = 0

A. I believe the last equation is the correct one: -4.9t2 + 24.5t + 117.6 = 0

15.
Q. Solve the equation you chose in question 18 to determine when the ball will hit the ground. (HINT: If you don't get one of the answers listed for this question, then maybe you chose the wrong equation in #18. Use this opportunity to double check your work!)
t = 8 seconds

t = 4 seconds

t = 3 seconds

t = -3 seconds

The ball will never reach the ground.

A. I believe the third one is correct: t = 3 as in it will take 3 seconds for the ball to hit the ground.

16.
Q. Using the same equation, determine when the ball is at a height of 49 meters.

A. It will take 7 seconds for the ball to reach 49 meters.

#183 Re: Help Me ! » Multiplying and Dividing Polynomials » 2012-12-10 14:10:14

Oh alright! And the: 8x^2/4x would become 2x?

#184 Re: Help Me ! » Multiplying and Dividing Polynomials » 2012-12-09 23:57:09

3.
(4x^5 + 8x^2) / 4x

4x^5/4x + 8x^2/4x

For first equation:
4x goes into 4x 1 time so it will be 1^5

For second equation:
4x goes into 8x 2 times so it will be 2^2.

It will then be:
1^5 + 2^2

I hope I did this one correct.

Number 4 seems to be pretty complicated. I'm not too sure how to start it!

#185 Re: Help Me ! » Multiplying and Dividing Polynomials » 2012-12-09 21:06:42

So that would be set as:

21x^3/7 + 14x/7

For the first fraction:
7 goes into 21x 3x times I suppose. This will leave it with 3x^3?

For the second fraction:
7 goes into 14x 2x times.

The problem then becomes:

3x^3 + 2x

They are not entirely alike so you can't add them. Therefor that is the answer I believe?

#186 Re: Help Me ! » Multiplying and Dividing Polynomials » 2012-12-08 22:03:56

So you don't reduce them?

So then:
3x/3x + 6x/3x

For first fraction:
3x goes into 3x 1 time.

For second fraction:
3x goes into 6x 2 times.

Then it becomes 1 + 2 and of course the answer is 3.

#187 Re: Help Me ! » Multiplying and Dividing Polynomials » 2012-12-08 18:51:18

Sorry I haven't answered sooner. I have been busy with other matters.

To finish from there. If I'm correct, we have to reduce it.
So:

3x/3x + 6x/3x

Reducing:
1x/1x + 3x/1x

For first fraction:
1x goes into 1x one time so it will become just x.

For second fraction:
1x goes into 3x three times so it will become 3x.

So the problem becomes x + 3x. So the answer is 4x?

#188 Help Me ! » Multiplying and Dividing Polynomials » 2012-12-05 23:56:10

demha
Replies: 25

I am having a hard time understanding how to do these correctly. Could someone please help explain how to do them step by step?


1. (3x + 6x)/3x


2. (21x3 + 14x) / 7


3. (4x5 + 8x2) / 4x


4. [2(x – 1) + 3(x – 1)] / (x – 1)


5. [3(2x – 3) – x(2x – 3)] / (2x - 3)


6. [x2(5x + 6) – 3(5x + 6)] / (5x + 6)


7. [3x2(2x - 3) + 27x3(2x - 3)] / [3x(3 – 2x)]

#190 Help Me ! » Applications of Systems of Linear Equations » 2012-12-01 23:15:26

demha
Replies: 3

For my studies. I'm not to sure how to do them. Especially number 2.


1. You want to borrow three rock CDs from your friend. She loves math puzzles and she always makes you solve one before you can borrow her stuff. Here’s the puzzle: Before you borrow three CDs, she will have 39 CDs. She will have half as many country CDs as rock CDs, and one-fourth as many soundtracks as country CDs. How many of each type of CD does she have after you borrow three rock CDs?


2. After watching hours of those home improvement shows, you decide you want to paint you bedroom. You don’t want to paint all four walls the same color (how boring!), but instead, you want to paint one wall a different color. The electric orange paint you’ve chosen for the "special" wall is more expensive, and you’re on a budget, so you need to know the area of the wall so you can buy the smallest amount of paint possible. You know that the height of the wall is half the length. You also know that the perimeter of the rectangular wall is 48 feet. What is the area of the wall?


A The area of the wall is 128 sq.ft.
BThe area of the wall is 116 sq. ft.
CThe area of the wall is 144 sq. ft.
DThe area of the wall is 140 sq. ft.
E The area of the wall is 416 sq. ft.
F The area of the wall is 136 sq. ft.

#191 Re: Help Me ! » Substitution and Elimination Method » 2012-11-26 22:27:50

Hello Bob! wink

So it was adding 4x & 4y to each side that would give me the correct answer. I see now. Then we have both similar equations then start the elimination to get 0 + 0 = 0 which in the end is 0 = 0: there is an infinate solution.

Thanks you very much Bob! I appreciate you taking your time helping me with this.

#192 Re: Help Me ! » Substitution and Elimination Method » 2012-11-26 19:57:38

Thanks for the welcome and super thanks for answering back! I knew that 1 - 9 & 10 - 18 were the same equations... but now I feel kind of stupid ._.
I should have known that I would get the same answers! =P

Here are the mistakes I redid:
4.
x + 2y = -4
4y = 3x + 12

x + 2y = -4
x = -4 - 2y

4y = 3(-4 - 2y) + 12
4y = (-12 – 6y) + 12
10y = -12 + 12
10y = 0
y = 0

x + 2y = -4
x + 2(0) = -4
x = -4

Answer: (4, 0)

---

7.
y = -2x + 1
y = x – 5

y = -2x + 1
(x – 5) = -2x + 1
      +5             +5
x = -2x + 6
+2x  +2x
3x = 6
x = 2

y = x – 5
y = (2) – 5
y = -3
Answer: (2, -3)

8.
y = (1/2)x - 3
y = (3/2)x – 1

y = (1/2)x – 3
(3/2)x – 1 = (1/2)x – 3
              +1                +1
(3/2)x = (1/2)x – 4
(4/2)x = -4
2x = -4
x = -2

y = (3/2)x – 1
y = (3/2)-2 – 1
y = -3 – 1
y = -4
Answer: (-2, -4)

---

(Elimination Method)
10.
y = (2/3)x - 1
y = -x + 4

0 = (-5/3)x + 5
(- 5/3)x = 5
x = 3

y = 2/3x – 1
y = 2/3(3) – 1
y = 2 – 1
y = 1
Answer: (3, 1)

---

16.
y = -2x + 1
y = x – 5

0 = 3x + 6
3x = 6
x = 2

y = x – 5
y = 2 – 5
y = -3
Answer: (2, -3)

17.
y = (1/2)x - 3
y = (3/2)x – 1

0 = -x – 2
x = -2

y = (3/2)x – 1
y = (3/2)-2 – 1
y = -3 – 1
y = -4
Answer: (-2, -4)

---

Now for number 18, I understand I am supposed to get the same answer as number 9 which is a "word solution." The answer should be infinate (0 = 0 for example). I have tried a few times but I just can't seem to get it. Would you mind helping me? Here is the equation yet again:

(Elimination Method)
18.
x + y = 2
4y = -4x + 8

#193 Help Me ! » Substitution and Elimination Method » 2012-11-25 23:20:39

demha
Replies: 5

This is for my school work. I am working online. Before I send it in I would like to know if it's all correct.

Solving Systems of Equations Using Substitution Method and Elimination Method:
SUBSTITUTION METHOD
1.
y = (2/3)x - 1
y = -x + 4

y = (2/3) - 1
(-x + 4) = (2/3)x – 1
(-1 1/2)x + 4 = -1
(-1 1/2)x = -5
x = 3

y = -x + 4
y = -3 + 4
y = 1
Answer: (3, 1)

2.
x + y = 0
3x + y = -4

x + y = 0
-x        -x
y = 0 – x

3x + (0 – x) = -4
2x + 0 = -4
2x = -4
x = -2

x + y = 0
(-2) + y = 0
+2           +2
y = 0 + 2
y = 2
Answer: (-2, 2)

3.
4x + 3y = -15
y = x + 2

4x + 3(x + 2) = -15
4x + (3x + 6) = -15
7x + 6 = -15
       -6     -6
7x = -21
x = -3

y = x + 2
y = (-3) + 2
y = -1
Answer: (-3, -1)

4.
x + 2y = -4
4y = 3x + 12

x + 2y = -4
x = -4 - 2y

4y = 3(-4 - 2y) + 12
4y = (-12 – 6y) + 12
10y = -12 + 12
10y = 0
y = 0

4y = 3x + 12
4(0) = 3x + 12
3x = 12
x = 4
Answer: (4, 0)

5.
y = 2x
x + y = 3
x + y = 3
x + (2x) = 3
3x = 3
x = 1

y = 2x
y = 2(1)
y = 2
Answer: (1, 2)

6.
x = 3 - 3y
x + 3y = -6

x + 3y = -6
(3 – 3y) + 3y = -6
3 + 0 = -6
3 = -6
Answer: There is no solution.

7.
y = -2x + 1
y = x – 5

y = -2x + 1
(x – 5) = -2x + 1
      +5             +5
x = -2x + 6
+2x  +2x
3x = 6
x = 2

y = x – 5
y = (2) – 5
y = 3
Answer: (2, 3)

8.
y = (1/2)x - 3
y = (3/2)x – 1

y = (1/2)x – 3
(3/2)x – 1 = (1/2)x – 3
              +1                +1
(3/2)x = (1/2)x – 4
(4/2)x = -4
2x = -4
x = -2

y = (3/2)x – 1
y = (3/2)-2 – 1
y = -3 – 1
y = 4
Answer: (-2, 4)

9.
x + y = 2
4y = -4x + 8

x + y = 2
y = 2 – x

4y = -4x + 8
4(2 – x) = -4x + 8
8 – 4x = -4x + 8
    +4x    +4x
8 – 8x = 8
-8          -8
8x = 0
0 = 0
Answer: This equation has infinite solutions.


ELIMINATION METHOD
10.
y = (2/3)x - 1
y = -x + 4

0 = (-5/3)x + 5
(-5/3)x = 5
x = -3

y = -x + 4
y = 3 + 4
     -3   -3
y = 1
Answer: (-3, 1)

11.
x + y = 0
3x + y = -4

2x + 0 = -4
2x = -4
x = -2

x + y = 0
-2 + y = 0
+2         +2
y = 2
Answer: (-2, 2)

12.
4x + 3y = -15
y = x + 2

y = x + 2 (times equation by 3)
3y = 3x + 6

4x + 3y = -15

7x = -21
x = -3

y = x + 2
y = -3 + 2
y = -1
Answer: (-3, -1)

13.
x + 2y = -4
4y = 3x + 12

x + 2y = -4 (times equation by 3)
3x + 6y = -12
4y = 3x + 12
3x - 4y = 3x – 3x + 12
4y – 3x = 12
3x + 6y = -12
3x + 6y + 4y – 3x = -12  + 12
6y + 4y = -12 + 12
10y = 0
y = 0

x + 2y = -4
x + 2(0) = -4
x = -4
Answer: (-4, 0)

14.
y = 2x
x + y = 3

x + y = 3 (times equation by 2)
2x + 2y = 6

y = 2x
y – 2x  = 2x – 2x
y – 2x = 0

y – 2x = 0
2x + 2y = 6
2x + 2y + y – 2x = 6 + 0
2y + y = 6 + 0
3y = 6
y = 2

y = 2x
(2) = 2x
2x = 2
x = 1
Answer: (1, 2)

15.
x = 3 - 3y
x + 3y = -6

x + 3y = 3 – 3y + 3y
x + 3y = 3

x + 3y = 3
x + 3y = -6

Answer: Both equations are the same, but both equal two different things. Therefore the answer is no solution.

16.
y = -2x + 1
y = x – 5

0 = -x + -4
x = -4

y = -2x + 1
y = -2(-4) + 1
y = 8 + 1
y = 9
Answer: (-4, 9)

17.
y = (1/2)x - 3
y = (3/2)x – 1

0 = -x – 4
x = 2

y = x – 5
y = (-2) – 5
y = 7
Answer: (2, -3)

18.
x + y = 2
4y = -4x + 8

4y = -4x + 8
4y – 4y = -4x – 4y + 8
-4x – 4y = 8

x + y = 2 (times equation by 4)
4x + 4y = 8

-4x + 4y = 8
4x + 4y = 8
8y = 16
y = 2

x + y = 2
x + (2) = 2
     -2     -2
x = 0
Answer: (0, 2)

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