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you have a choice
you can use the quadratic formula or you can complete the square
you pick

actually there is a third choice, namely to cheat

how do i do that, any of it ?

do you know the quadratic formula?

i need a refresher.

lets solve
\[x^2+4x+6=0\] by completing the square
ready?

sorry i was working it out xD

ok take your time

so far I got |dw:1438546685018:dw|

\[4^2-4\times 6=-8\]

\[x=\frac{-4\pm\sqrt{-8}}{2}\]

then
\[\sqrt{-8}=\sqrt{4\times 2\times (-1)}=\sqrt{4}\sqrt{2}\sqrt{-1}=2\sqrt{2}i\]

giving you
\[\frac{-4\pm2\sqrt{2}i}{2}\] divide each term by \(2\) and get
\[-2\pm\sqrt{2}i\]

it would have been easier to complete the square
there would have been no denominator

is that supposed to be the answer ?

lol yes it is "supposed to be" but in fact they wrote
\[-2\pm i\sqrt{2}\]

xD are you able to help me out with more ?

sure

lets complete the square for this one, unless you want to use the formula again

\[x^2+4x=1=0\] subtract 1 from both sides get
\[x^2+4x=-1\]

alright

then half of 4 is 2, and 2 squared is 4 go right to
\[(x+2)^2=-1+4\] or
\[(x+2)^2=3\]

take the square root of both sides, don't forget the \(\pm\) get
\[x+2=\pm\sqrt3\]

subtract 2 from both sides to solve for x, get
\[x=-2\pm\sqrt3\] done

looks like option C in your choices

xD I thought I wa supposed to do it lol

this is not a quadratic i guess

\[\frac{x}{x-4}=\frac{4}{x}\] right?

no that is not right
\[\frac{6}{x-4}=\frac{4}{x}\] hows that?

the x in the numerator is a 6

yea let me try xD

i will wait

x=-8?

yes

would it be extraneous?

how do i tell if its extraneous or non-extraneous?

in this case it does not, so it is not extraneous

oh now i get it

well actually
\[-\frac{1}{2}=-\frac{1}{2}\] but you get the idea

you up for some more xD? Or you tired out?

go ahead

What is the axis of symmetry for f(x) = 2x2 + 8x + 8?
x = −2
x = −4
x = 4
x = 2

oops
axis of symmetry of \(y=ax^2+bx+c\) is
\[-\frac{b}{2a}\]

in your case \(a=2,b=8\)

let me know when you get
\[x=-2\]

what is c?

nothing?

c in your case is 8 but it does not figure in to the axis of symmetry
it is just
\[x=-\frac{b}{2a}\]

so 2?

no there is a minus sign in the formula

oh so x=-2?

\[x=-\frac{8}{2\times 2}\\
x=-2\]

you need the slope first
you know how to find it?

y=mx+b?

what is the standard form again ?

standard form will be one of your first two choices
\[ax+by=c \]

im not understanding how i would use it

i really appreciate you helping me <3

\[\huge (b^m)^n=b^{m\times n}\] is what you need for this one, i.e.multiply the exponents

yw

so what would the b be?

multiply
\[\frac{2}{5}\times \frac{5}{6}\]

in this case \(b\) is \(2\)

do i cross multiply?

\[\huge (2^{\frac{2}{5}})^{\frac{5}{6}}\]
\[\huge 2^{\frac{2}{5}\times \frac{5}{6}}\]
hell no

multiply just means multiply

so\[\frac{ 10 }{ 30 }\]

\[\frac{2}{5}\times \frac{5}{6}\] we can cancel stuff too

the five

\[\frac{2}{\cancel{5}}\times \frac{\cancel{5}}{6}=\frac{2}{6}\] is a start

then you can cancel the 2 as well

or you could cancel here
\[\frac{1\cancel{0}}{3\cancel{0}}=\frac{1}{3}\]

either way you get \(\frac{1}{3}\) and also you do NOT cross multiply

ok gotcha cx makes more sense now

whew
final answer
\[\huge 2^{\frac{1}{3}}\]

lol

wait wouldnt it be x1/3

\[\frac{12z^2-7z-12}{3z^2+2z-8}\] factor and cancel

hold on let me go back and look

yeah sorry
\[\huge x^{\frac{1}{3}}\] the base was x, not 2

ahh gotcha

i dont understand how to use that xD

A?

click on the link to the second one and see how i wrote it in wolfram

yeah A

questions are all over the map
must be an on line course ...

no

yea it is XD

and oh i thought it was

one answer is
\[x^{\frac{4}{7}}\]

another is
\[\sqrt[7]{x^4}\]

the power is four, not the coefficient

\[x\times x\times x\times x=x^4\]not \(4x\)

thes are the answers

\[\huge \sqrt[7]{x^4}=x^{\frac{4}{7}}\]

make that a B

cheat like crazy for this one

http://www.wolframalpha.com/input/?i=+6x3+%2B+25x2+-+24x+%2B+5%3D0

\[\{-5,\frac{1}{2},\frac{1}{3}\}\]

so A?

otherwise you have to figure out that it factors as \[(3 x-1) (2 x-1) (x+5)\]

yeah A

Use the graph below for this question:
What is the average rate of change from x = −1 to x = −2?

\[5-7=-2\]

although the question is confusing since it says "from -1 to -2" which is backwards

maybe they want you to say 2, i am not clear on that
the question is ill posed

not uncommon for an on line class btw

sorry i was talking to my brother

thank you so much!!!

yw any time!