wrb said:
Hi Tutors:
No one is doing my homework. Im trying to figure out how to do it myself. The problem is, I'm not clear on your explanations. Please help me and stop being rude. I love your website. Thanks for all of your support. Mr Ohlston, can you explain things in a matter in which it could be understood.
Now, with all that said, can you please answer this question nicely. Show the Area=72 and Perimeter = 44
Thank you
If you are trying to figure out how to do it yourself, then I would think you would have TRIED something.
You've posted a bunch of questions, with not one stitch of work shown. So...how can we tell where you need help? We can give better explanations if we see what you have tried on your own.
In all of your posts, I've seen NOTHING that indicates any effort on your part. I don't think it is at all "rude" to ask you to show some effort.
Here's a couple of hints:
Area of a rectangle = length * width
If you have a rectangle with area of 72 (square units??????), then
length * width = 72
Can you find two numbers which multiply to 72?
And the perimeter of a rectangle is found using this formula:
P = 2*length + 2*width
Take the numbers you've found as possibilities for the length and width to get an area of 72. See if those numbers will satisfy the perimeter relationship...that is, that the perimeter is 44.
Here's one possibility. Suppose the width of your rectangle is 2. Then the width would be 36, since the area is to be 72. (2*36 = 72)
Now, if the length is 36, and the width is 2, what is the perimeter? Perimeter = 2*length + 2*width. If you have a length of 36 and a width of 2, then
perimeter = 2*36 + 2*2
or,
perimeter = 72 + 4
perimeter = 76
But that can't be right! The perimeter is supposed to be 44! Try a different pair of numbers which give you an area of 72.
Since you're posting these problems in the Arithmetic section of these forums, I assume you are not using things like calculus.