simplify x^2-9 over (x+3)(x+1)
Please help:)
Hint: Cancel factors of 1 \[\frac{(x + 3)(x - 3)}{(x + 3)(x + 1)}\]
I have no idea what to do next:(
if it's the exact same thing over the exact same thing it equals one. If it's one there's no need to write it.
so the answer is one right?
You can only cancel the parts that are the same not the whole thing.
\[\frac{\cancel{(x + 3)}(x - 3)}{\cancel{(x + 3)}(x + 1)}\]
I'm sorry I still don't get it.... is it (x-3) over (x+1)
yes
Lol I'm right
yup
You need to get it. It takes time to type these up :P
I know but i don't know how the steps work.
There's a lot for you to learn.
@Hero can you help with my question
what is your question?
The management of the UNICO department store has decided to enclose a 800 ft^2 area outside the building for displaying potted plants and flowers. One side will be formed by the external wall of the store, two sides will be constructed of pine boards, and the fourth side will be made of galvanized steel fencing. If the pine board fencing costs $6/running foot and the steel fencing costs $3/running foot, determine the dimensions of the enclosure that can be erected at minimum cost.
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so the equation would be 12x+6y right? I can only have one variable. but after that I'd find the derivative & critical values right? then what?
800 = xy P = 12x + 3y
Remember, we don't count the building wall
so P(800) = (12x+3y)+xy?
No bro
x = 800/y
what do i do since idk the lengths?
My teacher won't let us have two variables so the y is confusing me
Yes, exactly, so you substitute x = 800/y into the second equation bro
or you can substitute y = 800/x
P = 12x + 3(800/x) P = 12x + 2400/x
Now find P'(x)
P'(x)=12-2400/x^2?
Now set P'(x) = 0 and solve for x
2400/x^2 = 12 2400 = 12x^2 12x^2 - 2400 = 0 12(x^2 - 200) = 0 x^2 - 200 = 0 x^2 = 200 x = + or - sqrt(200) Distance can't be negative so x = sqrt(200) Good luck finding the rest
Thanx @Hero I'm sure you think I'm stupid by now but thanx a lot.
No bro, you're not stupid
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