#1
alright well heres the deal. this is how bad it is. in algebra II, usually pretty easy class for me. we had a group/open note test. i know sounds ridiculously easy, right? well i am probably in the worst group of 4 in the class. so i wrote some probs down on a peice of paper and we finish the test tomorow,and i am asking the pit for help on a few problems. many thanks for your answers. sorry if they are hard to readd with exponents and such...

anything in bold is an exponent. sorry i know its hard to read, it would probably be easier if you wrote it down on paper for yourself

11. simplify

(7a-3b4) (-3a4b-5)4


15. if f(x) = 6x+3 and g(x) = 5x-2, what restrictions are on the domain of (f/g)(x)?


any help is greatly appreciated.
thanks!
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#2
15 first, then I'll do 11.

your composite function is:

6x + 3
----------
5x - 2

So that function is undefined when the denominator is 0, or when g(x) = 0.

5x - 2 = 0
x = 2/5
#4
for restrictions, just remember that you cannot divide by 0. So anything that will make the denominator 0 is out.
#5
11. -84ab-1

15. x cannot equal 2/5
Between the velvet lies, there's a truth as hard as steel.
The vision never dies, life's a neverending wheel.
#6
Quote by demoniacfashion
Whyy has no one posted stupid comments about rape and cumming blood?

Fail.
#7
11.

I'm just going to pull the 4 out and stick it back in at the end

(7a - 3b^4) x (-3(a^4)b - 5)

-21(a^5)b - 35a 7a times the second term in parentheses.
9(a^4)(b^5) + 15b^4 3b^4 times the second term in parentheses.

So you get:
-21(a^5)b - 35a + 9(a^4)(b^5) + 15b^4

And you just multiply all the coefficients by 4.
#8
coool. thanks alot tigerking.

and edbert too for replying...

edit: and jewmasat...
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#9
11)
(7a - 3b^4) x (-3a^4 b^-5) x 4
=(7a - 3b^4) x (-12a^4 b^-5)
=(7a)(-12a^4 b^-5) + (-3b^4)(-12a^4 b^-5)
=(-84a^5 b^-5) + (36a^4 b^-1)

thats about as simple as it can go. you can factor out a 12 if you really wanted to. hope that helps
#10
Hmm, valvetronix, me, and edbert all posted different answers for 11 based on how we interpreted the exponents. Do you have a graphing calculator? I'd plug it into that and use the expand function for the correct answer. Or try to find an online utility that does that.
#11
Quote by tigerking615
11.

I'm just going to pull the 4 out and stick it back in at the end

(7a - 3b^4) x (-3(a^4)b - 5)

-21(a^5)b - 35a 7a times the second term in parentheses.
9(a^4)(b^5) + 15b^4 3b^4 times the second term in parentheses.

So you get:
-21(a^5)b - 35a + 9(a^4)(b^5) + 15b^4

And you just multiply all the coefficients by 4.


(7a - 3b^4) x (-3(a^4)b - 5) <---- this line is incorrect (at least it looks this way)

I thought the second term was (-3(a^4)(b^-5))... unless I read the question wrong
#12
yea i have one... not on me though. i just wrote down all the answers. ill use those to help me figure it out tomorow before class.
thanks alot everyone
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#13
Quote by tigerking615
Hmm, valvetronix, me, and edbert all posted different answers for 11 based on how we interpreted the exponents. Do you have a graphing calculator? I'd plug it into that and use the expand function for the correct answer. Or try to find an online utility that does that.


I think edbert is trying to add the last 2 terms together, which you cannot do, since they the 'a' and 'b' have different exponents
#14
Quote by ValvetronixRox
(7a - 3b^4) x (-3(a^4)b - 5) <---- this line is incorrect (at least it looks this way)

I thought the second term was (-3(a^4)(b^-5))... unless I read the question wrong



yaaa your write with the second term i beleive. just dont forget the exponent 4 outside the parenthesis
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