Woah! I did it!
It's been quite a few years since I've hadto do any geometry or algebra. I got a question from ChaCha asking to find the length and width of an area of 5000ft^2.
I had a question like this earlier and I tried to read a web page with instructions. The instrucions on the web page confused me. This time, I found a web page with problems like it and put it as the reference, but I worked out the problem - not sure where I was going at first. I did the work until I got a solution for the length and the width.
When I reached my solution, I was surprised and very happy that I had done the work to get the correct answer!
Here's what I ended up sending (people get ChaCha answers in text messages).
A=LW and 2L=W, so 5000=A and 5000=LW So 5000/L=W, now we know that 2*(5000/L)=L, so 10,000/L=L, multiply by L and 10,000=L^2. So 100=L. Now, 5000=100W So W=50. ChaCha has all your SOLUTIONS!
P.S. Did I mention I like the blockquote tag?
I had a question like this earlier and I tried to read a web page with instructions. The instrucions on the web page confused me. This time, I found a web page with problems like it and put it as the reference, but I worked out the problem - not sure where I was going at first. I did the work until I got a solution for the length and the width.
When I reached my solution, I was surprised and very happy that I had done the work to get the correct answer!
Here's what I ended up sending (people get ChaCha answers in text messages).
A=LW and 2L=W, so 5000=A and 5000=LW So 5000/L=W, now we know that 2*(5000/L)=L, so 10,000/L=L, multiply by L and 10,000=L^2. So 100=L. Now, 5000=100W So W=50. ChaCha has all your SOLUTIONS!
P.S. Did I mention I like the blockquote tag?