Recent content by Jenninifer

I got to the first step, but how do you get to the second step?

I've been using the formula lim x-->a f(x)-f(a) -------- x-a I haven't had any problems until I was asked to find the slope of the tangent at the general point whose x-coordinate is a. How do I do it with an a instead of a point? I'm trying to find it for...

How about: 3-3 + 3-4 _____________ 3-5

-4a^2?

I don't get either. But i just need how to simplify it.

Yes that's it! I still don't get it though.

Sorrry I don't know how!

Homework Statement 4a^4b^3 -a^3 ---------- x ------- a^5b^6 -(b^2) Homework Equations The Attempt at a Solution (4a^4b^3)(1a^3) = 4-a^7b^-3 --------------- a^5-b^4 = $-a^2b^-1 WHERE DID I GO WRONG!

okay, i thought the book was wrong but just thought i'd make sure. and there is no way 7 and 3 can make 12.

I don't think that's what it's asking for. It's just simplifying it. I wrote what it says the answer is. But i got like -y squared -3x+5y-4

(8y squared + 5y -7) - ( 9y squared + 3x -3) This is my friends, and I'm trying to hlep him on it. And the answer in the book is -y squared +8y -12 and we can't figure out how to get it.

yes so 2m = 60, divide both by 2 and you get 30, WOW! i cannot believe myself.

i know it's 5m and 3m right?

okay well what do you get when you multipy it? do you get 15m over 3 and 15m over 5 equals 60?

I tried that already, i multiplied everything by 15. but the answer for m i got was 20 and when you substitute it it doesn't work.