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 Posted: Sun Feb 25, 2007 3:46 pm
My friend screwed me and never came over to help me correct my test. PLEASE HELP ACE crying This is the back of my test which I did the worst on. http://i12.tinypic.com/308jkeg.pngIf you could just rework any of the problems and tell me what I did wrong, I will be so happy. I really don't want to fail this class.
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Posted: Sun Feb 25, 2007 3:48 pm
gonk egga's in her 2 year of algebra
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Posted: Sun Feb 25, 2007 3:50 pm
EggaSponge gonk egga's in her 2 year of algebra I bet you could do it though. I know it's not hard but I can't get it.
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Posted: Sun Feb 25, 2007 3:56 pm
Goddess Ace EggaSponge gonk egga's in her 2 year of algebra I bet you could do it though. I know it's not hard but I can't get it. i am so sorry ace crying i am struggleing with alg2 right now i dont think i can help you and i read some of the problems and i at least tried to figure them out and couldnt
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Posted: Sun Feb 25, 2007 3:57 pm
The best I can do is try to get Jill to help though she's not great with graphs.
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Posted: Sun Feb 25, 2007 4:00 pm
I should be able to help Ace, just give me a second
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Posted: Sun Feb 25, 2007 4:00 pm
[Tsukasa567] I should be able to help Ace, just give me a second Thank-you heart
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Posted: Sun Feb 25, 2007 4:03 pm
*stares in confusion till' head asplodes all over the walls* burning_eyes
I know!! write "CHUCK NORRIS" in really big letters everywhere!! instant A+++!!! 3nodding rofl xd xd
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Posted: Sun Feb 25, 2007 4:09 pm
Number 10 should be pretty easy to explain.
Okay, so you are trying to find the inverse of that problem right? So, change it from.
Y= sqr(5-x)+7
To
X=sqr(5-y)+7
From here you will want to get the y by itself, so first subtract 7 from both sides making this:
x-7=sqr(5-y)
Next, perform the opposite of a square root by squaring both sides, you would then get:
(x-7)^2=5-y
You can foil that out, or leave it as it is, not sure what your teacher wants. anyway, next subract 5 from both sides making:
(x-7)^2-5=-y
or
x^2-14x+44=-y
Final step, divide by -1 and change y back to f(x) which means you get:
f(x)=-1(x-7)^2+5
or
f(x)= -x^2+14x-44
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Posted: Sun Feb 25, 2007 4:12 pm
All right, I'll write that out.
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Posted: Sun Feb 25, 2007 4:14 pm
Sorry. Artanus knows not the ways of calculus sad
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Posted: Sun Feb 25, 2007 4:15 pm
[Tsukasa567] Number 10 should be pretty easy to explain. Okay, so you are trying to find the inverse of that problem right? So, change it from. Y= sqr(5-x)+7 To X=sqr(5-y)+7 From here you will want to get the y by itself, so first subtract 7 from both sides making this: x-7=sqr(5-y) Next, perform the opposite of a square root by squaring both sides, you would then get: (x-7)^2=5-y You can foil that out, or leave it as it is, not sure what your teacher wants. anyway, next subract 5 from both sides making: (x-7)^2-5=-y or x^2-14x+44=-y Final step, divide by -1 and change y back to f(x) which means you get: f(x)=-1(x-7)^2+5 or f(x)= -x^2+14x-44 All I can say is WOW eek and then clapping for you comes to mind.
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Posted: Sun Feb 25, 2007 4:16 pm
Next to go to number 9
For part c, all I have to tell you is to remember that it is always x= don't just put the numbers, but put x= in front of them, otherwise you got that right.
Part d, I'm pretty sure that the maximum value of f is actually (4,3) or (4,4) I can't really tell, the graph goes above the point at which your scale stops.
Part f, look for where it is going from low to high, so it is (1,-3) to (4,whatever the y is)
Part g, look at what the y value is when x=4 and multiply that by 2
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Posted: Sun Feb 25, 2007 4:20 pm
Question 12, the only real thing you seem to have messed up on is the x is less than or equal to zero part.
Basically what it is saying is anytime that the x value for the function is less than or equal to zero, then you square that and it is your y value.
so for example, x=0 would be 0 since 0^2 is still zero. Then you move to x=-1, square that and you have 1, so your point would be (-1,1). -2 would be 4 so the point would be (-2,4) and so on and so forth into negative infinaty.
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Posted: Sun Feb 25, 2007 4:22 pm
Wow, you teach better than my own teacher.
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