# Functions--> homework , absolutly stuck :S

• Apr 11th 2008, 03:37 PM
mike_302
Functions--> homework , absolutly stuck :S
Hi,
Normally I wouldn't do this, but I recieved a tonne of htis functions stuff. I'm actually very efficient in completing the square to find zeros, but on these questions I recieved, I had 4 questions that, my answer wasn't/isn't going to match up with the ones in the book. There is one question on each page that I marked with a star (on the page) and I uploaded to photobucket. I'll copy out the questions here:

15. The base of a rectangle has a perimeter of 300m. The area of the base is 4400m^2. What are the dimensions.

17. The base of a triangle is 2 cm more than the height. The area of the triangle is 5 cm^2. Find the base to teh nearest 10th of a centimeter.

23. Two positive integers are in a ratio of 1:3. If their sum is added to their product, the result is 224. Find the integers.

24. Two whole numbers differ by 3. The sum of their squares is 89. What are the numbers?

As you can see, these questions are simple word problems, solved by completing the square. Let me know if you have any questions about my methods. I really hope to hear back PLENTY on this because it is simple stuff which seems to elude me (the answers at least.. The idea, I get) . I didn't post the answers in the book because I figured you guys could just tell me if I did anything wrong, but if you want the answer for a particular question, let me know. I'll be available a LOT this weekend, but the sooner hte better. Thanks in advance! (btw, I'm not asking one person to answer them all, so if you dont have time to go through all 4, could you just check one or two?)

http://i160.photobucket.com/albums/t...athhelp004.jpg
http://i160.photobucket.com/albums/t...athhelp003.jpg
http://i160.photobucket.com/albums/t...athhelp002.jpg
http://i160.photobucket.com/albums/t...athhelp001.jpg
• Apr 11th 2008, 04:54 PM
mike_302
scratch number 17. In waiting, I received a reply from a friend concerning that one and discovered my silly error in the first line... (1/2b*h)/2 ... yes... To err is human (Doh).

but no word on any of the others! Anyone have anything for me?
• Apr 11th 2008, 05:09 PM
mr fantastic
Quote:

Originally Posted by mike_302
[snip]

15. The base of a rectangle has a perimeter of 300m. The area of the base is 4400m^2. What are the dimensions. Mr F says: 2x + 2y = 300 => x + y = 150 => y = 150 - x. Therefore A = xy => 4400 = x(150 - x). Solve for x.

17. The base of a triangle is 2 cm more than the height. The area of the triangle is 5 cm^2. Find the base to teh nearest 10th of a centimeter. Mr F says: A = 1/2 (Base)(Height). Solve 5 = 1/2 (h + 2)(h) for h.

23. Two positive integers are in a ratio of 1:3. If their sum is added to their product, the result is 224. Find the integers. Mr F says: x/y = 1/3 => y = 3x. Therefore x + y + xy = 224 => x + 3x + x(3x) = 224. Solve for x.

24. Two whole numbers differ by 3. The sum of their squares is 89. What are the numbers? Mr F says: x - y = 3 => x = 3 + y. Therefore x^2 + y^2 = 89 => (3 + y)^2 + y^2 = 89. Solve for y.

[snip]

I leave the many details to you.
• Apr 11th 2008, 05:35 PM
mike_302
OK, a question as to number 15. I followed along with completing the square after getting the two equations you pointed out: l*w=4400 and l+w=150 ... The differnce between yours and mine is that I used 2l+2w=300 . I easily see how you reduced those, but I don't understand why that makes a difference. I assume that both statements are true, so as long as I solved for one variable or another correctly, and used it in my equation, I would think it would work out to the same thing. It obviously didn't ... But why not?
• Apr 11th 2008, 05:42 PM
mr fantastic
Quote:

Originally Posted by mike_302
OK, a question as to number 15. I followed along with completing the square after getting the two equations you pointed out: l*w=4400 and l+w=150 ... The differnce between yours and mine is that I used 2l+2w=300 . I easily see how you reduced those, but I don't understand why that makes a difference. I assume that both statements are true, so as long as I solved for one variable or another correctly, and used it in my equation, I would think it would work out to the same thing. It obviously didn't ... But why not?

It makes no difference to the answer (but obviously makes the calculation easier). I get x = 110 or x = 40. x = 40 => y = 110 ..... If you're not getting the same answers for l and w, then you're making a mistake.

Show you working.
• Apr 11th 2008, 05:45 PM
mike_302
http://i160.photobucket.com/albums/t...athhelp002.jpg
I posted my work to that question :S perhaps you can look at my work and point out where I went wrong. In that page I had used 2l+2w=300 ... (I finished the answer after the first reply I got back here, and got the correct answer with l+w=150 , but I want to figure out where I went wrong, and I'm not seeing it :S)
• Apr 11th 2008, 05:53 PM
mr fantastic
Quote:

Originally Posted by mike_302
http://i160.photobucket.com/albums/t...athhelp002.jpg
I posted my work to that question :S perhaps you can look at my work and point out where I went wrong. In that page I had used 2l+2w=300 ... (I finished the answer after the first reply I got back here, and got the correct answer with l+w=150 , but I want to figure out where I went wrong, and I'm not seeing it :S)

The line next to the big star has been got by expanding incorrectly. It should be

$-2(w - 75)^2 {\color{red}+} 11250 = 4400$.

By the way, in your subsequent calculations you are taking the square root of a negative number - NO REAL SOLUTION!

A lot of your problems are probably the result of carelessness. So ..... BE CAREFUL!
• Apr 11th 2008, 06:06 PM
mike_302
It really is a lot of carelessnees :( I'm disappointed in myself. But yes, I realize the whole root of a negative problem, I just realized after a step or two. I was pretty far into a heavy load of repeated questions (15 word problems involving the same type of math) so I was going through extremely quick after a while.