lokiisnotdead

and also, this is probably a really dumb question, but the inverse of 4 is 2  and the inverse of 5 is 1 in this situation right? sorry if this was a dumb question, I've just never learned this stuff before.

thank you so much geno! I'm starting to understand this now!
 

I just have 1 question, for number 1, why does 1+0=0 and why does 2+0=0? Isn't 1+0=1 and 2+0=2?

hi guest!

so we can start by raising \(m-\frac{1}{m}=3\) to the 2nd power.

\((m-\frac{1}{m})^2=(3)^2\)

\(m^2 + 2m\cdot \frac{1}{m} + \frac{1}{m^2} = 9\)

\(m^2 -2+ \frac{1}{m^2} = 9\)

\(m^2+\frac{1}{m^2} = 11\)

now let's square this again

\((m^2+\frac{1}{m^2} )^2= 121\)

\((m^2)^2 + 2m^2\cdot\frac{1}{m^2} + \left(\frac{1}{m^2}\right)^{\!2} = 121\)

\(m^4+2+\frac{1}{m^4}=121\)

\(m^4+\frac{1}{m^4}=\boxed{119}\)

i got 18... i'm  not sure if im right though

hmm I think a better approach to this problem is to use complementary counting.

There are \(\dbinom{6}{3}=20\) ways to choose 3 points to make a triangle. However 2  of these ways are just straight lines, so \(20-12=\boxed{18} \text{ triangles. }\)

I'm not sure if this is right, so please let me know if it's correct or not! 

:)

Since it's perpendicular, the slop would be the negative reciprocal of -2, which would be 1/2.

It intersects the point (8,2) which means that 2=1/2(8)+b.

2=1/2(8)+b

2=4+b

b=-2

So, the new equation is \(\boxed{y=\dfrac{1}{2}x-2}\)

Nevermind, I do remember how to solve this. 

First let's cube \(x-\frac{1}{x}=2\)

\((x-\frac{1}{x})^3=2^3\)

\(x^3-3x+\frac{3}{x}-\frac{1}{x^3}=8\)

\(x^3-\frac{1}{x^3}=8+3(x-\frac{1}{x})\)

\(x^3-\frac{1}{x^3}=8+3(2)\)

\(x^3-\frac{1}{x^3}=\boxed{14}\)

I can't remember how to solve this problem, but I'm pretty sure the answer is 14.

yeah that's what i initially thought too, but the answer is actually 5 - sqrt(15)·i or  5 + sqrt(15)·i

\(\boxed{z=3+2i} \text{ and } \boxed{z=4-i}\)

yeaaa i'm an army too lol

\(4 \frac{3}{10} + 3 \frac{2}{5}+ \frac{1}{3} + \frac{1}{2}=\boxed{\frac{128}{15}} \text{ or } \boxed{8\frac{8}{15}} \)

\(ab + b^2 - ac -bc\)

\(= b(a+b)b(a+b)\)

\(=\boxed{ (a+b)(b-c)}\)

lmao wish i could help u but im going into 8th grade :(

good luck in high school tho!

umm actually i think the gcd of 36 and 104 is 4.

gwenspooner did the prime factorization right, but what she missed is that 3^2 is not a factor of 104. so, the gcd is 2^2=4.