+8262 I don't understand why the answer is is listed like that below. In the homework, the answer is $$\frac{b-7}{a^3}$$
Please help. I don't want to register, so please help. Thank you.
Question:
$$\dfrac{a^2b-7a^2}{a^5}$$
Possible answers:
$$\boxed{\dfrac{b-7}{a^3}},\mbox{\;or\;}\boxed{a^2b-\dfrac{7}{a^3}}$$
+129899 Note, DS, that we have the law of exponents that says an / am = a n-m
So.....look at the fraction....we have
7 * ( a2 / a5) and using this law, we have
7 * (a 2-5 ) = 7 * a-3
And remember that a-n = 1 / an
So.......
7 * a-3 = 7 * ( 1 / a3) = (7 / a3 )
So...our answer should be
a2b - (7/a3)
Does that help ???
+129899 Note, DS, that we have the law of exponents that says an / am = a n-m
So.....look at the fraction....we have
7 * ( a2 / a5) and using this law, we have
7 * (a 2-5 ) = 7 * a-3
And remember that a-n = 1 / an
So.......
7 * a-3 = 7 * ( 1 / a3) = (7 / a3 )
So...our answer should be
a2b - (7/a3)
Does that help ???
+8262 It does, but it is confusing. And the answer is not there in the homework. I will shoot the photo, and you would see that the answer isn't there, and it will be my proof and evidence that the answer is not there. So, do I shoot the photo?
+129899 Isn't my answer the same thing as your second "possible" answer ???
I'm confused !!! ????????
+8262 Yes it is. And it is incorrect, I guess. I will shoot the photo.
+8262 I may need help on the ones of subtracting. Can you teach me how to do it?
+129899 OK.....if we have
[a2b - 7a2] / a5 we can factor out an a2 at the top and we have
[(a2) (b - 7)] / a5 =
[b - 7 ] / a3
+8262 I have a question. In the first step, why do you cancel out $$a^2?$$
+118687 Hi Dragon 
$$a^2b-7\frac{a^2}{a^5}\\\\
a^2b-\frac{7}{1}\times\frac{a^2}{a^5}\\\\
a^2b-\frac{7\times a^2}{1\times a^5}\\\\
a^2b-\frac{7a^2}{a^5}\\\\
a^2b-\frac{7aa}{aaaaa}\\\\
a^2b-\frac{7\not{a}\not{a}}{\not{a}\not{a}aaa}\qquad \mbox{Note: aa cancels out}\\\\
a^2b-\frac{7}{aaa}\\\\
a^2b-\frac{7}{a^3}\\\\$$
+118687 Dragon, you wrote that before you even had time to think about what I had written!!!
Now go back and think about it properly!
+8262 May someone explain how you canceled out aa???
And can someone help me figure out to do in the following problem please?:
$$\dfrac{6a^2-30a+36}{4a-12}$$
How do I start off, and can you give an example of how to do it?
+129899 [ 6a2 -30a + 36 ] / [4a - 12 ] factor out the common factor of 6 at the top and 4 at the bottom
(6/4) [a2 - 5a + 6 ] / [a - 3] factoring the trinomial, we have
(6/4) [(a -3)(a-2)] / (a-3) "canceling the "(a-3)'s " and reducing (6/4) we have
(3/2)(a - 2) or (3a - 6) / 2 ...whichever you prefer !!!
+8262 Oh! I get the answer now! Melody, how do you know it is my first MathWay question? This is my third, but I didn't post it here...
+8262 Can you guys help on the following?
$$\dfrac{3a^3(16-a^2)}{12a^6(a^2-9a+20)}$$
I got to the part:
$$\dfrac{1(16-aa)}{4a^3(a^2-9a+20)}$$
I don't know what to do next. Can you pleas tell me what to do next?
+8262 May anyone help me here!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!??!?!?!??!!?!
+129899 3a3(16- a2) / [12a6 (a2 - 9a + 20)] factoring everything, we have
3a3 (4 - a)(4+a) /[ 12a6 (a - 4)(a-5)] =
-(1/4a3 )(a-4)(a+4)/[(a-4)(a-5)] = (we can factor a "-" out of (4 - a) = - (a - 4))
-(1/4a3)(a+4)/(a-5) = (5-a)(4+a)/(4a3)
+8262 Thank you! I need to understand what to do for other similar questions like that one. Can you please tell or explain what to do in questions like these?
thanks. I would give you hundreds and hundreds of points CPhill.
+129899 I don't have any "magic" formula for doing these, DS......it's mainly just factoring and "cancelling"......plus a few algebraic "tricks".......
+118687 Much of this is WAY OVER your head Dragon. Your teacher cannot seriously expect you to do it!!
+8262 I know. The teacher said, " try your best. I will tell you how on Friday"
so, I try to make the best work I can. Ok. I know. It is not my level, but thanks for helping.
Only help on T.
