# Need help. Test on Monday!

• Oct 4th 2008, 05:53 PM
Chinnie15
Need help. Test on Monday!
Hello everyone.

I am in desperate need of help. I am in college algebra and just can NOT get modeling down, and I have a test on Monday! I am completely lost, and realize I should have gotten help before now but now I have no choice but to learn it all today and tomorrow. It's just the set up, I know how to do the problems. Is there any where besides my text that I can get help with these application problems? They're killing me, and I'm really worried I'm gonna fail this first test.

I do have a direct question though. I've been working through these the best I can, and just can not solve this one no matter what I do.

A woman cycles 8 mi/h faster than she runs. Every morning she cycles 4 mi and runs 2.5 mi, for a total of one hour of exercise. How fast does she run?

I know that her running speed is x, and that her cycling speed is 8+x, but I still can't get anywhere. Please help if you can. :(
• Oct 4th 2008, 06:02 PM
galactus
We know d=rt. We are given that the total time is 1 hour.

Since d=rt, then t=d/r.

Let x=running time, as you stated.

$\frac{4}{8+x}+\frac{2.5}{x}=1$
• Oct 4th 2008, 06:05 PM
Jonboy
this is kinda confusing. it all boils down to: $d = rt$

which can be made: $t = \frac{d}{r}$

For running, you can make: $\frac{2.5}{x} = t_{r}$

$t_{r}$ just means time of running

For biking, you can make: $\frac{4}{x + 8} = t_{c}$

Now $t_{r} + t_{c} = 1$ so now you have:

$\frac{2.5}{x} + \frac{4}{x + 8} = 1$

Just solve for x and you're on your way.
• Oct 4th 2008, 06:39 PM
Chinnie15
Thank you so much guys! :) I'm going to be doing a load of review/studying tonight, and if I hit any road blocks I'll post them here.

Here is one more I just hit.

A parcel of land is 6ft longer than it is wide. Each diagonal from one corner to the opposite corner is 174 ft long. What are the dimensions of the parcel?

x= width
6+x= length

I get that, but I have no idea to do with the 174ft diagonal line, and how to set it up. It's like this with all of them, I can get some of it, it's just getting the rest that's been the problem. But once I understand it, I can easily apply it to the other problems.
• Oct 4th 2008, 06:53 PM
icemanfan
Quote:

Originally Posted by Chinnie15
A parcel of land is 6ft longer than it is wide. Each diagonal from one corner to the opposite corner is 174 ft long. What are the dimensions of the parcel?

x= width
6+x= length

The diagonal is the hypotenuse of a right triangle with sides x and 6+x. Therefore, its length is $174 = \sqrt{x^2 + (6+x)^2}$.

Square both sides and solve for x.
• Oct 4th 2008, 07:01 PM
Jonboy
it's better to start another thread if you have another problem. if someone sees all these replies and you put your next question on here it won't get answered as fast.