Question

algebra 2 11th please help. thx!

Answer 1

#25 is the 1st one i think.

Answer 2

thanks. better than nothing

Answer 3

i see ?????s is that right?

Answer 4

but you should get -2,1, if not we can walk through it.

Answer 5

Is it still showing that? Ill retype it. gimme a sec, tried copy and pasting to fix an math error.

Answer 6

yeah i have it. thanks
25?

Answer 7

#24
Use the rational number test


\[x^4+5x^3+7x^2-3x-10=0\]

find the factors of 10 so

\[1,2,5,10\]

so that means \[\pm 1,2,5,10 \over 1\]

over 1 because the first coeffecient is 1

so that means all your possible zeros are at

\[-10,-5,-2,-1,1,2,5,10\]

just plug each value into the original function and you should get the answer, I got -2,1 so try it and if you get something else we'll figure it out

Answer 8

there it! thanks

Answer 9

yeah i have -2, 1

Answer 10

I'm working out #2, i have the answer gimme a sec to post proof.

Answer 11

#25 Find the zeroes of \[12x^2-64 = -x^4\]

add \[-x^4\] to both sides

so then you have

\[x^4-12x-64=0\]



then factor

\[x^2+12y-64=0\]

into \[(y-4)(y+16)=0\]

Then you have
\[x-4 =0 \] and \[x+16 =0\]

add 4 on the x-4 and you get

\[x^2 =4\]
same for the +16

\[x^2=-16\]

take the square of both sides and you get

\[\pm 2 \]
and \[\pm4i\]

Answer 12

Sorry about the wait, equation writer didn't want to work with me.

Answer 13

thank you!

Answer 14

did you get #26 yet?

Answer 15

nope

Answer 16

#26

\[(a+b)^n = \sum_{k=0}^{n}\left(\begin{matrix}n \\ k\end{matrix}\right) a ^{n-k}b^k\]

thats the formula to use..do you understand how it works?

Answer 17

so its C?

Answer 18

Yes.

Answer 19

ok thank you so much! can you help me with more? i will open the next question on another post

Answer 20

As long as its not too much more.