asdf335

i am 313 now :)

use upload when you make the image

i think cphill told you there was no image last time

HOPE THIS HELPED!

ok here is my check: f(6)-f(5) should equal 3 according to you, 

according to me it should be -7.

plugging in f(6), it is 2+24-36 or -10, and f(5) is 2+20-25 or -3, so it is -7 and i am right.

hope this helped, and the answer is actually -h-6.

oh.... sorry

i have seen this before, i will show you.

also to everyone else who is confused about the operations, it is \(\frac{f(5 + h)-f(5)}{h}\), not something else.

now, we have to calculate the numerator. f(h+5) is 2+4h+20-h^2-10h-25, and f(5) is -3. now you have (-h^2-6h)/h for the whole fraction, so the answer is -h-6. 

someone check my work pleez

and hope this helped!

yes, if you move the -19y the the other side, it is like adding 19y to both sides, so the sign would change to cancel things.

then you would have the equation 7x=10+19y.

HOPE THIS HELPED!

lol lenny face

hey if you can, i can personally answer them and show you a bit about how to do them ;)

for the first question, you take out a \(25x^2y^6\) and take the square root, so you get \(5xy^3\), and then you still have a \(3x\) still in there, so first one's answer is 

\(\boxed{5xy^3\sqrt{3x}}\).

for 2nd question, subtract 2 from both sides, then square, to get \(2x-1=x^2-4x+4\).

now take away \(2x-1\) from both sides, to get \(x^2-6x+3\). now use quadratic formula, \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\), and plug in the values to get \(x = {-(-6) \pm \sqrt{(-6)^2-(4)(1)(3)} \over 2}\). im gonna let you do the rest, just for a tiny exercise 

for 3rd one, every solution will make one of those terms 0, so the answers are \({1 \over 2}, 8, \) and \(-1\).

the last one, plug everything in to get \(4^4-2(4)^3+5(4)^2-20(4)-4\).

and that is your second little exercise 

HOPE THIS HELPED!

you have x^4+2x^2y^2+y^4+2x^2y^2-y^4 on the left side, which makes x^4+4x^2y^2. on right side, it is x^4+4x^2y^2. these are both equal.

HOPE THIS HELPED!