lokiisnotdead

well this isn't a good way to solve the problem and I'm sure there's many other smarter ways to solve it, but
 

\(531340_6 = 5(6^5)+3(6^4)+6^3+3(6^2)+4(6)= 38880+3888+ 216+108+24 = 43116_{10}\).

the remainder of that when divided by 210 is 66.

thank you so much! i understand it a lot better now!!!

wow tysm! that really helped! :)

I can't really tell what fraction that is... please fix the latex! :)

I think what s/he meant was

If m is an integer is \(m^3\cdot 3^{k+3}-m^2\cdot 3^{2k+3}+m\cdot 3^{k+2}-1\) an integer?

thank you! yeah I forgot about adding a-b=8 and a+b=8 together. thank you so much for the work you put into the clear solution! :)

thank you so much!!! i see what i did wrong now. thank you for the great and clear solution!!! :)

hi guest!
 

we can start off by finding the y-coordinate which is (0,3),so that means that b=3.

next we can find the slope by using \(\dfrac{y_2-y_1}{x_2-x_1}\), and we can use (-1,0) and (0,3).

This would equal \(\dfrac{3-0}{0-(-1)}=\dfrac{3}{1}=3\).

so the answer is \(\boxed{y=3x+3}.\)

also is there a way to delete questions?

yeah i tried that

792=8*9*11

so c=6

then, 1+3+a+b+9+6=multiple of 9

so, a+b=8 or a+b =17

for it to be divisible by 11, 1-3+a-b+4-5+6= multiple of 11, so 3+a-b= multiple of 11, which means that a-b=8 or a-b = -3.

but then i tried to plug in the equations and i got x=7 and y=10 :/

can someone please tell me what I'm doing wrong?

hello there!

hi!

i wish i could answer this, but unfortunately i dont know python :(

i think you would have much better luck on Stack Overflow, because this site is mostly for math and science 

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 \(x-\frac{1}{x}=2\)

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

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

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

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

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