[(l,k)]

Originally Posted by DJWildefire

Alright. I'd start with a diagram.
Attachment 59482

The barn wall doesn't count toward the 40 meters of fencing. Therefore, only one length of the pen needs fencing but both widths do. Thus, 2x+x+x=40 m. x, the width, equals 10 meters. Therefore, 2x, the length, equals 20 meters.

thanks man. the wording confused me into thinking i had to include the barn wall but i understand now. thank you very much

[(l,k)]

Originally Posted by DJWildefire

Alright. I'd start with a diagram.
Attachment 59482

The barn wall doesn't count toward the 40 meters of fencing. Therefore, only one length of the pen needs fencing but both widths do. Thus, 2x+x+x=40 m. x, the width, equals 10 meters. Therefore, 2x, the length, equals 20 meters.

here's another. "an artist is designing a rectangular stained glass window in such a way that the length of the window is 3 inches less than twice its width. what are the dimensions of the window if its area is 740 square inches?" i honestly dont know why im having a problem with these, they seem like the easiest ones at first. it wasnt covered in our notes though so thats probably it.

[DJWildefire]

Originally Posted by (l,k)

here's another. "an artist is designing a rectangular stained glass window in such a way that the length of the window is 3 inches less than twice its width. what are the dimensions of the window if its area is 740 square inches?" i honestly dont know why im having a problem with these, they seem like the easiest ones at first. it wasnt covered in our notes though so thats probably it.

Probably best to start with a (poorly drawn) diagram again.

(2x-3)(x)=740
2x^2-3x-740=0
2x^2-40x+37x-740=0
2x(x-20)+37(x-20)=0
(x-20)(2x+37)=0
x=20 (can't be negative)
x, the width, is 20 inches. Thus, 2x-3, the length, is 37 inches.
20 in x 37 in

[(l,k)]

Originally Posted by DJWildefire

Probably best to start with a (poorly drawn) diagram again.
Attachment 59492
(2x-3)(x)=740
2x^2-3x-740=0
2x^2-40x+37x-740=0
2x(x-20)+37(x-20)=0
(x-20)(2x+37)=0
x=20 (can't be negative)
x, the width, is 20 inches. Thus, 2x-3, the length, is 37 inches.
20 in x 37 in

2x^2-3x-740=0
2x^2-40x+37x-740=0


did you factor that first line to get the second?

[DJWildefire]

Originally Posted by (l,k)

2x^2-3x-740=0
2x^2-40x+37x-740=0


did you factor that first line to get the second?

Well you either have to factor 2x^2-3x-740 or use the quadratic formula. The way I was taught to factor polynomials when coefficient a is greater than one is by multiplying a and c together and finding the factors that add to b, then substituting those in as coefficients in place of b, and factoring by grouping. e.g. a=2, b=-3, c=-740 --> 2(-740) = -1480 --> (-40)(37) = -1480, -40+37=-3 --> -3x=-40x+37x
So from the first line to the second, I didn't factor, I just rearranged so that I'd be able to factor by grouping in the next step.

I'm not sure how common it is for teachers to teach factoring that way; I know that some advocate the use of a guess and check method when a>1 and with coefficients that large it'd be annoying to guess and check. So some people would probably just use the quadratic equation.

[(l,k)]

Originally Posted by DJWildefire

Well you either have to factor 2x^2-3x-740 or use the quadratic formula. The way I was taught to factor polynomials when coefficient a is greater than one is by multiplying a and c together and finding the factors that add to b, then substituting those in as coefficients in place of b, and factoring by grouping. e.g. a=2, b=-3, c=-740 --> 2(-740) = -1480 --> (-40)(37) = -1480, -40+37=-3 --> -3x=-40x+37x
So from the first line to the second, I didn't factor, I just rearranged so that I'd be able to factor by grouping in the next step.

I'm not sure how common it is for teachers to teach factoring that way; I know that some advocate the use of a guess and check method when a>1 and with coefficients that large it'd be annoying to guess and check. So some people would probably just use the quadratic equation.

yeah she talked about guess and check but she never presented us with a problem like this, and we havent learned the quadratic formula just yet, so i guess i was doomed lol. thanks a bunch though

[DJWildefire]

Originally Posted by (l,k)

yeah she talked about guess and check but she never presented us with a problem like this, and we havent learned the quadratic formula just yet, so i guess i was doomed lol. thanks a bunch though

That sucks; math problems are really difficult to figure out if you haven't been taught the concepts you need for the problem yet haha. Hopefully she'll start giving you problems that actually test what you she's taught you. No problem.

[xhmnimx]

I have a couple. I'm terrible at this stuff, so any help would be greatly appreciated.

1. <1 and <2 are supplementary angles. m<1 = x-12, and m<2 = x+100. Find the measure of each angle.

2. On a number line, p has a coordinate -33 and q has coordinate 46. Find pq.

[DJWildefire]

Originally Posted by xhmnimx

I have a couple. I'm terrible at this stuff, so any help would be greatly appreciated.

1. <1 and <2 are supplementary angles. m<1 = x-12, and m<2 = x+100. Find the measure of each angle.

2. On a number line, p has a coordinate -33 and q has coordinate 46. Find pq.

Supplementary angles add to 180 degrees. So m<1+m<2=180. Therefore,
x-12+x+100=180
2x-12=80
2x=92
x=46
Then substitute x back into m<1 = x-12 and m<2 = x+100.
m<1 = 46-12 = 34 degrees
m<2 = 46+100 = 146 degrees

As for your second question, I'm not exactly sure what that question means. If you're finding pq, that's just the product of p and q, so I'd assume it'd just be (-33)(46) which equals -1518. But maybe you're using a notation I'm unfamiliar with.

[xhmnimx]

Originally Posted by DJWildefire

Supplementary angles add to 180 degrees. So m<1+m<2=180. Therefore,
x-12+x+100=180
2x-12=80
2x=92
x=46
Then substitute x back into m<1 = x-12 and m<2 = x+100.
m<1 = 46-12 = 34 degrees
m<2 = 46+100 = 146 degrees

As for your second question, I'm not exactly sure what that question means. If you're finding pq, that's just the product of p and q, so I'd assume it'd just be (-33)(46) which equals -1518. But maybe you're using a notation I'm unfamiliar with.

Thanks a ton. I really appreciate it. I'm surprised at how active this thread is, hah, but I'm glad I found it. As far as the second question, I'm not sure either. This class is extremely hard for me to pay attention to because of the idiots surrounding me never stop talking. But thanks again, man.

[tonyC4L]

Anyone want to take my dynamical systems midterm for me tomorrow? Or teach me about Lyapunov functions and limit cycles better than my crappy book and lecture notes? I will gladly blow you.

[DJWildefire]

I've got a test on Taylor and Maclaurin Series tomorrow. Don't think it'll be too bad but you never know.

[Apsira]

The Best Colleges in India provides more information about education to students and it is help to students for how to choose best colleges you have to visit this site http://allaboutedu.com/best-colleges-in-india

[DJWildefire]

Originally Posted by tonyC4L

...

Hey, sorry to bug you but I'm struggling with one aspect of Taylor series manipulation and I'm wondering if you could help?
Here's an example of a problem:
f(x)= {(cosx-1)/x^2 for x =/= 0, -1/2 for x=0}
Use the Taylor series for cos x about x=0 to write the first three nonzero terms and the general term of the Taylor series for f about x=0.

I know the cosine series by memory but I'm having trouble manipulating the series properly.
Thanks so much!

edit: nevermind, I was subtracting 1 from each term rather than from the whole cosine series. feelin' idiotic.

[tonyC4L]

Originally Posted by DJWildefire

Hey, sorry to bug you but I'm struggling with one aspect of Taylor series manipulation and I'm wondering if you could help?
Here's an example of a problem:
f(x)= {(cosx-1)/x^2 for x =/= 0, -1/2 for x=0}
Use the Taylor series for cos x about x=0 to write the first three nonzero terms and the general term of the Taylor series for f about x=0.

I know the cosine series by memory but I'm having trouble manipulating the series properly.
Thanks so much!

edit: nevermind, I was subtracting 1 from each term rather than from the whole cosine series. feelin' idiotic.

Hehe glad you got it sorted out