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Willam
Aug 26th, 2009, 07:34 AM
is there anyone good @ math around that can help me with some problems please?

moby
Aug 26th, 2009, 08:00 AM
What problems?

Willam
Aug 26th, 2009, 08:27 AM
ill post it tomorrow, is almost 4am and i decided to give up.

TheFifthAvocado
Aug 26th, 2009, 02:18 PM
i love math :tears: hurrry up!

Sally Struthers
Aug 26th, 2009, 02:22 PM
i love math :tears: hurrry up!

right on.

http://matttarr.com/nerdPower.jpg

nerd power! :lol:

Kart
Aug 26th, 2009, 02:47 PM
Tzar, I'd take Sal up on her offer quick. She doesn't do charity often :p.

InsideOut.
Aug 26th, 2009, 02:48 PM
I'd love to help. Post your question NOW. :(

Willam
Aug 26th, 2009, 03:25 PM
ok so i have 20 excersises to solve, only 10 left though.

ill post 3 for now :). lol.

i have to clarify it has been over a year since i graduated from high school im trying to find my way back in college so.

1. A fram of uniform width borders a painting that is 20 inches long and 13 inches high. If the area of the framed picture is 368 square inches, find the width of the border.

What i did was. 20x13 = 260

368-260 = 108

108/20= 5.4
108/13= 8.3

8.3-5.4= 2.9 inches, but im not sure if its right.

2. Determine the constant that should be added to the binomial:

X^2 + 1/6^X

so that it becomes a perfect square trinomial. then write and factor the trinomial.

3. A box with a square base and no top is made from a square piece of cardboard by cutting 3-inch squares from each corner and folding up the sides. What length of side must be the original cardboard square have if the volume of thebox is 675 cubic inches.

4. 5X^4 - 45X^2 = 0.

5. Solve the ecuation.

X-2048-32.SQUARE ROOT of X = 0.

please post procedures.

god i feel so stupid. i know how to do this but i forgot everything i learned on highschool

Meteor Shower
Aug 26th, 2009, 03:49 PM
Hmm, Let me try the first question..

I think for the first its you got 108 pixels for the border like you calculated.
Its uniform so with the variable X that is the border you weight, you need to "circle" the picture, so its (2*height + 2*length) * X = 108

height=13, length=20. So its should be 108/66?

Noctis
Aug 26th, 2009, 03:53 PM
let see

Meteor Shower
Aug 26th, 2009, 03:55 PM
Hmm, Let me try the first question..

I think for the first its you got 108 pixels for the border like you calculated.
Its uniform so with the variable X that is the border you weight, you need to "circle" the picture, so its (2*height + 2*length) * X = 108

height=13, length=20. So its should be 108/66?

Oh, BTW, with the calculation 2*height + 2*length you have the four corner pixels repeated twice, so really it should be 62 and not 66. So the answer should be 108/62 I guess.

Bart
Aug 26th, 2009, 04:05 PM
4. 5X^4 - 45X^2 = 0.

5x(x-9) => x=0, x=3 or x=-3

Kworb
Aug 26th, 2009, 04:35 PM
1.

(20+x)(13+x) = 368
260 + 33x + x = 368
x + 33x = 108
x = 3

border width = x/2 = 1.5

Sally Struthers
Aug 26th, 2009, 04:37 PM
is #2 written correctly?

I don't think it is. I'll do it how I think it should be written

x^2 +1/6x

complete the square

take half of 1/6 and square it

(1/6)/2 = 1/12
(1/12)^2 = 1/144

so:

x^2 +1/6x +1/144

factored: (x+1/12)(x+1/12)

Kworb
Aug 26th, 2009, 04:41 PM
2.

x + 1/6x + a = (x + b)

x + 1/6x + a = x + 2bx + b

b = 1/12

a = 1/144 = constant that should be added
x + 1/6x + 1/144 = perfect square trinomial
(x + 1/12) = factored into binomials

Kworb
Aug 26th, 2009, 04:46 PM
3. volume is 3*(x-6)*(x-6) = 675

x - 12x + 36 = 225
x - 12x - 189 = 0
(x-21)(x+9) = 0 --> x=21

InsideOut.
Aug 26th, 2009, 04:49 PM
Damn I missed the first few questions :o More, Tzar?

moby
Aug 26th, 2009, 04:52 PM
1. A fram of uniform width borders a painting that is 20 inches long and 13 inches high. If the area of the framed picture is 368 square inches, find the width of the border.Let width be of border be x inches
Then length of framed picture = 20 + 2x
Height of picture = 13 + 2x
(20 + 2x)(13 + 2x) = 368
4x^2 + 66x + 260 = 368
2x^2 + 33x - 54 =0
(2x - 3)(x + 18) = 0
x = 1.5 (take the positive value)

2. Determine the constant that should be added to the binomial: X^2 + 1/6X

so that it becomes a perfect square trinomial. then write and factor the trinomial.Since the quadratic is a square, it is of the form (X + a)^2 = X^2 + 2a + a^2
By comparison, a = 1/12, so a^2 = 1/144 is the constant term.

3. A box with a square base and no top is made from a square piece of cardboard by cutting 3-inch squares from each corner and folding up the sides. What length of side must be the original cardboard square have if the volume of thebox is 675 cubic inches.Height of box = 3 inches
Base area of box = 675/3 = 225 = 15^2
Base of box has side 15 inches.
Length of original cardboard = 15 + 3 + 3 = 21 inches

4. 5X^4 - 45X^2 = 0.5x^2(x^2 - 9) = 0
5x^2(x + 3)(x - 3) = 0
x = 0 (repeated root), -3 or 3

5. Solve the ecuation.

X-2048-32.SQUARE ROOT of X = 0.Let u = square root of x
We have a quadratic: u^2 - 32u -2048 = 0
Note that 2048 = 2^11, 32 = 2^5
(u - 2^6)(u + 2^5) = 0
u = 2^6 or -2^5
square root of x = 2^6 or -2^5 (take the positive value, the principal square root is positive)
x = 2^12

TheFifthAvocado
Aug 26th, 2009, 05:15 PM
loves it! :hearts:

drake3781
Aug 26th, 2009, 05:42 PM
:rolls:

Meteor Shower
Aug 26th, 2009, 05:48 PM
Nicole, Kaia and Alexandra walk into a Nike Store. They are buying a tennis brain for $30. Each womman pays $10. The Store manager Maria S realizes that the brain only costs $25! She grunts and sends her assistant Maria K to the store with $5. Each woman gets $1 back, and Bohomova steals $2.

So lets review - each ended up paying $9. Considering the $2 steal, that's only $29 total.
Where's the missing dollar?

gentenaire
Aug 26th, 2009, 05:52 PM
They paid 3*9= 27 dollars, while it was only 25. Maria stole the extra 2.

Meteor Shower
Aug 26th, 2009, 05:59 PM
Yep :p only it was Kat, not Maria K

Another one:
Elena starts a serving practice.
She has a probablity of 60% making a DF at every ball.

How many balls Lena is going to serve in average until she hits her first DF?

Kart
Aug 26th, 2009, 06:13 PM
Yep :p only it was Kat, not Maria K

Another one:
Elena starts a serving practice.
She has a probablity of 60% making a DF at every ball.

How many balls Lena is going to serve in average until she hits her first DF?
Depends on the occasion.

If it's a grand slam final, she'll start with a double fault.

TheFifthAvocado
Aug 26th, 2009, 06:32 PM
More WTA math:

Serena wants to build a rectangular enclosure around her Big Macs. She only has a $900 Marbella first round loss check to spend on the fence and wants the largest size for her money. She plans to build the security fence along the moat around her mansion, so she does not have to put a fence on that side. The side of the fence parallel to the fence will cost $5 per foot to build, whereas the sides perpendicular to the river will cost $3 per foot. How many games will she drop on the way to the U.S. Open championship??

There's an answer to the actual question too:lol:

Apoleb
Aug 26th, 2009, 06:33 PM
Yep :p only it was Kat, not Maria K

Another one:
Elena starts a serving practice.
She has a probablity of 60% making a DF at every ball.

How many balls Lena is going to serve in average until she hits her first DF?

:lol: Loves it.

Geometric distribution so the mean is 1/p = 1.6 ~ 2.

Meteor Shower
Aug 26th, 2009, 06:43 PM
:lol: Loves it.

Geometric distribution so the mean is 1/p = 1.6 ~ 2.


correct :p :yeah:

Meteor Shower
Aug 26th, 2009, 06:44 PM
Depends on the occasion.

If it's a grand slam final, she'll start with a double fault.

:rolls:

The Crow
Aug 26th, 2009, 06:44 PM
:lol: Loves it.

Geometric distribution so the mean is 1/p = 1.6 ~ 2.

That's the double fault included, right? (shifted geometric). The average number of good serves before a DF would be 2/3 ((1-p)/p), if I'm not mistaken.

Meteor Shower
Aug 26th, 2009, 06:49 PM
That's the double fault included, right? (shifted geometric). The average number of good serves before a DF would be 2/3 ((1-p)/p), if I'm not mistaken.

Is that the calculation in case the probability is 60% for a single serve, and you have to get two out for a double?

Willam
Aug 26th, 2009, 06:57 PM
you want more, i can give u all my homework :)

Apoleb
Aug 26th, 2009, 07:00 PM
That's the double fault included, right? (shifted geometric). The average number of good serves before a DF would be 2/3 ((1-p)/p), if I'm not mistaken.

You're right. It's 2/3. Poor Lena.

moby
Aug 26th, 2009, 07:33 PM
More WTA math:

Serena wants to build a rectangular enclosure around her Big Macs. She only has a $900 Marbella first round loss check to spend on the fence and wants the largest size for her money. She plans to build the security fence along the moat around her mansion, so she does not have to put a fence on that side. The side of the fence parallel to the fence will cost $5 per foot to build, whereas the sides perpendicular to the river will cost $3 per foot. How many games will she drop on the way to the U.S. Open championship??

There's an answer to the actual question too:lol:43 games. :lol:

The actual question is an optimisation question with constraints, so we use Lagrange multipliers. If we have x as the length of the enclosure, and y the width, then the total cost is 5x + 6y.

So we set g(x,y) = 5x + 6y - 900 (spending all the money will give us the biggest fence) and we want to maximise f(x,y) = xy.

Now d/dx g(x,y) = c d/dy f(x,y), where c is the Lagrange multiplier, etc.
5 = cy
6 = cx
so y=5/c, x=6/c

5(6/c) + 6(5/c) = 900
60/c = 900
c = 2/30

so y = 75, x = 90
and xy = 6750 square feet is the maximum.

moby
Aug 26th, 2009, 07:42 PM
The Sabotage

Justine Henin is in the lockerroom chilling before the first Grand Slam final of her comeback. She has 12 identical-looking rackets, all of equal weight (because she's a stickler about things like swingweight), except... not really. You see, moments before the final starts, the locker room attendent delivers a mysterious note to her, scribbled in scraggly handwriting.

"Dear Christine, I have tampered with one of your rackets and changed its weight because I was board bored. Good luck!

Love,
Se SJW
Maria Sharapova"

Justine discovers a makeshift beam balance (duh!), but notices that the balance scale would probably disintegrate after 3 weighings. Help Justine figure out which is the rouge racket, and whether it is heavier or lighter!

Willam
Aug 26th, 2009, 09:00 PM
i could always post 6 more:p

Apoleb
Aug 26th, 2009, 09:39 PM
The Sabotage

Justine Henin is in the lockerroom chilling before the first Grand Slam final of her comeback. She has 12 identical-looking rackets, all of equal weight (because she's a stickler about things like swingweight), except... not really. You see, moments before the final starts, the locker room attendent delivers a mysterious note to her, scribbled in scraggly handwriting.

"Dear Christine, I have tampered with one of your rackets and changed its weight because I was board bored. Good luck!

Love,
Se SJW
Maria Sharapova"

Justine discovers a makeshift balance beam (duh!), but notices that the balance beam would probably disintegrate after 3 weighings. Help Justine figure out which is the rouge racket, and whether it is heavier or lighter!

What's a makeshift balance beam?

The Crow
Aug 26th, 2009, 09:50 PM
The Sabotage

Justine Henin is in the lockerroom chilling before the first Grand Slam final of her comeback. She has 12 identical-looking rackets, all of equal weight (because she's a stickler about things like swingweight), except... not really. You see, moments before the final starts, the locker room attendent delivers a mysterious note to her, scribbled in scraggly handwriting.

"Dear Christine, I have tampered with one of your rackets and changed its weight because I was board bored. Good luck!

Love,
Se SJW
Maria Sharapova"

Justine discovers a makeshift balance beam (duh!), but notices that the balance beam would probably disintegrate after 3 weighings. Help Justine figure out which is the rouge racket, and whether it is heavier or lighter!

:lol: good one! I've heard this one before, so I'm gonna leave it for others ;)

FORZA SARITA
Aug 26th, 2009, 09:52 PM
i feel so dumb reading this thread :spit::o

Noctis
Aug 26th, 2009, 09:58 PM
Rename this as homework thread :spit:
i am sure people need help aswell

moby
Aug 26th, 2009, 10:08 PM
What's a makeshift balance beam?Oops, sorry, it's actually called a beam balance. Or a balance scale.

http://www.hhs.gov/web/images/balance.png

Meteor Shower
Aug 26th, 2009, 10:10 PM
The Sabotage

Justine Henin is in the lockerroom chilling before the first Grand Slam final of her comeback. She has 12 identical-looking rackets, all of equal weight (because she's a stickler about things like swingweight), except... not really. You see, moments before the final starts, the locker room attendent delivers a mysterious note to her, scribbled in scraggly handwriting.

"Dear Christine, I have tampered with one of your rackets and changed its weight because I was board bored. Good luck!

Love,
Se SJW
Maria Sharapova"

Justine discovers a makeshift balance beam (duh!), but notices that the balance beam would probably disintegrate after 3 weighings. Help Justine figure out which is the rouge racket, and whether it is heavier or lighter!

I. take 2 groups of three.
if its not similar
II. take 1 new group of three (you know for sure that one is legit) and 1 already weighted
if its not similar (the already weight is the not-legit)
III. compare random two out of the three in the one the one that is not legit

if you get similar in any of those steps, just choose the other one.

Is this right? :o

Sally Struthers
Aug 26th, 2009, 11:34 PM
I have a math problem for everyone:

When is 3 < 0 ?


















solution:
when it's the 2008/2009 WTA season :haha:

Miss Amor
Aug 27th, 2009, 12:14 AM
I have a math problem for everyone:

When is 3 < 0 ?


















solution:
when it's the 2008/2009 WTA season :haha:


:rolls: :rolls:

moby
Aug 27th, 2009, 05:17 AM
I. take 2 groups of three.
if its not similar
II. take 1 new group of three (you know for sure that one is legit) and 1 already weighted
if its not similar (the already weight is the not-legit)
III. compare random two out of the three in the one the one that is not legit

if you get similar in any of those steps, just choose the other one.

Is this right? :oNo. Why?

Meteor Shower
Aug 27th, 2009, 07:26 AM
My logic was first to find a group that all of its rackets are legit, then to compare that group with the group that has the fake racket. Now I know if the fake racket is heavier or lighter based on the scale results (if the scale went up or down). Last thing I gotta do is find that racket, take two of the group of three. If they are equal, its the third. If they are different, the one that is lighter/heavier (depends what my result was in the second scale) is the fake one.

I realized my mistake now btw, so no need to explain why its wrong.