Please help me, simplify, leave in exponent form 5(rt)^9

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Please help me, simplify, leave in exponent form 5(rt)^9

Mathematics
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Expand, then simplify. Eventually you'll be able to do the simple way, but for now lets keep with the basics.
If you want to have someone check your answer please post the answer you got and ask for a check. Just posting the problem makes it appear as though you're not doing the work or understanding the material which makes people tend to not want to continue helping. If you show that you've done it and you just want to be sure you're correct people are more than happy to confirm.
so r*r*r*r*r*r*r*r*r t*t*t*t*t*t*t*t*t

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i still dont understand i get some of the problems but a few of them like this and the last I just cant figure out
It's ok. you're doing really well. \[5(rt)^9 = 5(rt)(rt)(rt)(rt)(rt)(rt)(rt)(rt)(rt)\] \[=5rrrrrrrrrttttttttt\] Right?
yes so 5r^9t^9?
yes.
yayyy, can you explain the FOIL method to me?
Sure. You know that a(b+c) = ab + ac Right?
yes so like (3x-6)(x-12) = 4x+18
That's jumping ahead, and not quite correct. Lets back up a step.
haha ok sorry
4(x + 3) = ?
4x+12?
Right. But we know that 4 is 3 + 1, so we should get the same thing if we rewrite it as (3+1)(x+3)
So lets take that whole first factor and multiply it by each of the terms in the second factor \[ \rightarrow (3+1)(x+3) = (3+1)x + (3+1)3\]
And then we distribute again.. \[(3+1)x+(3+1)3 = 3x + 1x + 9 + 3\]
\[ = 4x+12\]
sorry had to step away for a moment
=4x+12?
So now if we have something like: (a+b)(c+d) we can distribute: \[(a+b)(c+d) = (a+b)c + (a+b)d = ac + bc + ad + bd\]
But to make it easy to remember they prefer you to use FOIL which does the same thing, but in a different order.
+bd
i have like 5 sections and the first says use the foil and the rest just says find the product
(a+b)(c+d) They take the F irst two terms(a and c) O uter two terms(a and d) I nner two terms (b and c) L ast two terms (b and d) ac + ad + bc + bd Which is the same as what we had, but with a slightly different order.
oh okay im going to work a problem really quick can you check my answer?
yep
(3x-6)(x-12) 4x+3x-12-6x+18
Afraid not. (3x-6)(x-12) Multiply the F irst two terms (3x and x) 3x * x = ?
4x
Nope. 4x = 3x + x
3x^2
Correct. Now the I nner two terms (-6 and x) -6 * x = ?
-6x
Now the O uter two terms(-6 and -12) -6 * -12 = ?
72
Blah! Those are not the outer terms, they're the Last terms. Sorry.
Now the O uter two terms(3x and -12) 3x * -12 = ?
is it -36x? or does it say the same since they are not like terms?
it is -36x, just like -6 * x was -6x.
You can rearrange the factors of a product. a*b*c = a*c*b = b*c*a = c*a*b = ... etc
so now i have: 3x^2-6x^36x+72?
yes, except that it's -6x-36x, not -6x^36x but I think that was a typo.
yes sorry so now it is 33?
-6-36 = ?
-42
so -6x-36x = x(-6-36) = ?
x(-42)
So \[ 3x^2 - 6x - 36x - 72 = ?\]
i thought i had the answer, im never going to get through college :( 3x^2-x(-42)-72
You're really close. The only problem is you left an extra - there.
before 72?
We said that -6x-36x = x(-42) which we can re-arrange to be -42x So 3x^2 -6x -36x - 72 = 3x^2 -42x - 72
ohh man im never going to pass, im sorry you had to take your free time to help me. thank you so much im just having the hardest time
no you're doing great Think about it like this. If you have 4x's and you subtract 36x's you'll be at -32x's right?
right
everytime i think i get it a new problem comes up and im so lost again
so if you have -6x's and you take away 36x's you'll be at -42x's
That's ok. New problems should challenge you. Otherwise you'd just be doing things you already know and you wouldn't learn anything.
yes your right, im just frustrated if i cant get this how am i supposted to get the next two classes

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