I need math help ASAP

[Yet_Another_Boring_Marxist]

So I'm taking a business algebra test this tuesday and I've been trying to learn the material for the past week. I thought I was doing well, whenever I asked friends about problems I didn't know they explained them and it made sense but now I am trying to do all of those problems again and I realize that I can't learn math in this enviroment,. I already flunked the first exam (49%) and I REALLY need to get at least a B or an A on the next one. I've been studying since 8 and now its 1:12 in the morning but I haven't the slightest clue, can one of you help- me out? I have skype we can do math help over that please please please please please please

[bcbm]

could you post a sample problem or two?

[BIXX]

Post a sample problem, I'd be down to Skype if I know how to help.

[Yet_Another_Boring_Marxist]

Sorry for taking so long to respond, I'm working on a problem as we speak, when I am done with it I'll post some

[RedBen]

i can try, not great at algebra but i got the fundamentals... good luck anyway it goes. a sample would be helpful though

[Yet_Another_Boring_Marxist]

here it is


A rectangular plot of ground measuring 12 meters by 20 meters is surrounded by a sidewalk of a uniform width,. The area of the sidewalk is 68 square meters. find the width of the walk


This is what I tried:

(12-2x)(20-2x)=68

240-24x-40x+4x^2=68

4x^2-64x-240x=68

4(x^2-16x-76)=0

and when I try to factor it I have no luck


and I would love to skype

[RedBen]

here it is



This is what I tried:

(12-2x)(20-2x)=68

240-24x-40x+4x^2=68

4x^2-64x-240x=68

4(x^2-16x-76)=0

and when I try to factor it I have no luck


and I would love to skype

4 feet i believe. best of luck, i'm gonna bow out now before i mess you up.

[Yet_Another_Boring_Marxist]

4 feet i believe. best of luck, i'm gonna bow out now before i mess you up.

There are two answers. 8 by 14 inches. What I need to know is how to solve the problem

[bcbm]

well i found the solution (http://www.algebra.com/algebra/homework/word/geometry/Geometry_Word_Problems.faq.question.230898.html) online but i can't really explain it. i don't remember algebra at all, sorry. :(

[BIXX]

Hey, I'm figure out my skype, give me a minute.

I MIGHT be able to help.

[Yet_Another_Boring_Marxist]

ok its 2:33 AM where I am, I'm going to go to sleep but i'll be on tommorow

[BIXX]

ok its 2:33 AM where I am, I'm going to go to sleep but i'll be on tommorow

I think I figured out how to explain it, lets set up a time for tomorrow. Best of luck!

[Quail]

Hey, if you need any more help when you wake up I'll probably be around on facebook later so send me a message and I will get back to you.

As for the problem...

A rectangular plot of ground measuring 12 meters by 20 meters is surrounded by a sidewalk of a uniform width,. The area of the sidewalk is 68 square meters. find the width of the walk

The area of the sidewalk, given as 68 m squared, is equal to the area of the sidewalk and the plot of ground, minus the area of the plot of ground. The area of the plot of ground is 20*12=240. If we denote the width of the sidewalk by x, then the area of the sidewalk and the plot of ground is (20+2x)(12+2x).

Putting that together we get:
(20+2x)(12+2x)-68=240
(20+2x)(12+2x)-308=0
4x^2+64x-68=0 (multiply out and simplify)
x^2+16x-17=0 (divide by 4)
(x+8)^2-81=0 (complete the square)
(x+8)^2=81
x+8=9 (take square roots, but because this is a length it must be positive so just take the positive root)
x=1

Hope this is of some use.

[Czy]

A rectangular plot of ground measuring 12 meters by 20 meters is surrounded by a sidewalk of a uniform width,. The area of the sidewalk is 68 square meters. find the width of the walk

area of ground = 12* 20 = 240m^2
so ( 2x+12) ( 2x+ 20) - 240 = 68
or 4x^2 + 64x + 240 - 240 =68
or 4x^2 + 64 x - 68=0
or x^2 + 16x -17 =0
or x^2 + 17x - x - 17=0
or x( x+ 17) - 1 ( x+ 17)= 0
or ( x+ 17) ( x- 1) =0
so x= 1 & x=-17 ( invalid)
x= 1 meter =width of side walk

edit: PM me if you need any other maths help, man. Always ready to help

[Alonso Quijano]

here it is



This is what I tried:

(12-2x)(20-2x)=68

240-24x-40x+4x^2=68

4x^2-64x-240x=68

4(x^2-16x-76)=0

and when I try to factor it I have no luck


and I would love to skype
Quail and Czy already nailed it, but notice your mistake:

(12-2x)(20-2x)=68
What is this supposed to mean? What is the formula that this represents?
x is the width of the sidewalk. You tried to approach that given area (68), not looking at other stuff you know.

I've drawn it at first:
First, draw the rectangle, at supposedly 12 and 20. Now, add an outer bar - the sidewalk.

You have a rectangle within a rectangle. The outer "bar" is 68 square meters (given). The inner one is 240 (you can caclulate it - 12 times 20).

The big rectangle (all included)'s area is length times width.
The big rectangle's length: 20+2x
The big rectangle's width: 12+2x
(We know it's a uniform width, so we call it x)
Now, knowing that the sidewalk area is 68, and being able to calculate ourselves that the inner rectangle is 240 square meters, that leaves us with a big rectangle (original one+sidewalk) which is 68+240=308 square meters.

And to formulate:
(20+2x)(12+2x)=308

To summarise, also look for what you have, what you know, and what you don't.
What you have after drawing:
-A rectangle made of a smaller one plus a bar surrounding it
What you know:
-The area of the small one (from the facts)
-The area of the bar (given)
Therefore:
-The area of the entire rectangle

area=lengthXwidth

You have the area, you've defined both width and length and 12+2x and 20+2x, respectively.

Being good at math really is knowing how to utilise what you know, and what to search for. Your mistake was in the first equation. There were no calculation problems.

Post another sample, and explain what you see, what information you can gather, and how you approach it.

[Czy]

(20+2x)(12+2x)=308 ---------------- [308 = 68+240]
240+64x+4x^2=308
4(60+16x+x^2)=308
60+16x+x^2=77
(x+10)(x+6)=77 ------------------ [77 = 11*7]
so x+10 = 11
x = 1

here's another way of simplifying. Sometimes people prefer different methods so here's an additional one, might make it easier :)

[cyu]

We should run a math workshop out of this forum - sounds like it could be helpful, cool, and maybe even fun :wub:

[The Garbage Disposal Unit]

Hey YABM!
As someone else who often has difficulty dealing with numbers in the abstract, my feeling is that, generally, if you can draw a diagramme, do it. For me, having a practical, spacial representation often saves me from doing nonsense calculations.
In this case, two labeled squares makes a lot more sense to me than "(20+2x)(12+2x)"

[Sea]

(12-2x)(20-2x)=68Expand the "(12-2x)(20-2x)" out so you get a quadratic of the form ax^2+bx+c and schlop that 68 over to the other side so you get a quadratic equation of the form ax^2+bx+c=0. Check for a common divisor, if there is one factor it out. Complete the square, or re-factor and use the zero-factor property to solve, or whatever other method you want.


240-24x-40x+4x^2=68

4x^2-64x-240x=68

4(x^2-16x-76)=0These are just quadratic equations.Check for a GCD, if there is one factor it out. Get the 68 over to the left side on the first two. Complete the square.