power27

The divisor pairs of 6 are 1 and 6, and 2 and 3. Summing the factors of 6 excluding 6 gets 1+2+3=6, so no matter how many times you apply the function to 6, the answer will always be 6.

If there are n people, you would count the number of handshakes as nC2.

nC2= \(\frac{n(n-1)}{2\cdot1}\)

If there were 10 people, there would be (10*9)/2=45 handshakes. Thats not enough, so lets try a bigger number. 12 people: (12*11)/2=66 handshakes so there were 12 people.

There are 6 letters so 6! (6*5*4*3*2*1) ways, except there are two Zs and 6! counts the Zs as different so you need to divide by 2.

6!/2 = 360 ways

The rows of Pascal's triangle sum to powers of 2, so the elements of the nth row (the row with only one element in it being row 0) sum to 2ⁿ (there's a really cool counting proof for this but you probabally don't need to know that)

so f(n)=log(2ⁿ)

Put this into the fraction:

\(\frac{\log{2^{n}}}{\log{2}}\)

Pop that n out front from logarithm laws:

\(\frac{n\cdot\log{2}}{\log{2}}\)

The log(2)s cancel out and you are left with \(\frac{f(n)}{\log_{10}2}=\boxed{n}\)

I can usually avoid studying in most classes by paying attention and doing hw and stuff.

If that fails, then if the teacher gives you some list of things to know for the test thats usually a good place to start. I usually look over that and fill in stuff I don't know from notes and stuff. (this is also a great resourse for cramming right before the test, totally don't know that from experience)

When I need to learn a list of vocab words I use Quizlet. Also comming up with weird associations between words and definitions helps me remember things, like what a word sounds like or how its spelled. Even sitting there trying to come up with something to remember a word by helps.

It usually helps me to look at something, come up with a few words or short phrases about it, say it in my head/wisper it, say it w/o the paper a few times, then come back later and see if I can still remember it. This is good for knowing a concept or comparing/contrasting stuff, where you really only need to know the general idea, then be able to lengthen and add detail to it on the test.

Anyways my study habbits are atrocious, so take this with a grain of salt (or maybe a whole shaker full).

1. You got it good job!

2.

You are given a measure, either an equation or a degree measure, so these can be substituted into the exuation.

180=(136-2a)+(3a+39) so the answer is D

3. *measures the short side with fingers*

*puts it on the long side*

Looks like the short side is about half the long side so 4.2 or C

4. Thats correct, never heard of that method before but it seems like it works.

To find the smallest value, you need to find the vertex of the porabola. The vertex is the point that is the lowest on the porabola (or highest if a is negative). The x value produces the lowest (or highest) value of the porabola, so we only need to find one half of the point.

To find the x value, plug the coefficients into \({-b \over 2a}\)

\({-(-9) \over 2(1)}\)

=9/2

=4.5 will produce the lowest value of the porabola

Yeah I did oops

I was testing to see if you actually understood

26 total letters - I, O and Q = 23 letters you can use

10 total letters - 0 = 9 numbers you can use

There are 26 options for the first of the three letters, and 25 options for the second because it can be anything but the first letter, and 25 options for the third letter because it can't be the second letter but it can be the first letter

26*25*25=16250 ways to make the letters

Try to use the same logic I used to count the number of ways to make the numbers, and then multiply the number of ways to count the letters by the number of ways to count the numbers 

For the first problem, an equilateral triangle has sides that are all the same length. One of the sides of the triangle is on a side of the square, and all four sides of a square are the same length, so the sides of the triangle and the sides of the square are the same length. To find the side length of the square, first think about how you would find the area given a side... you would square the side. What number squared is 25?

To get the prerimiter of the square add up all the sides of the figure, so 3 sides of the square and 2 sides of the triangle.

I would help with the 2nd and 3rd but the images are blocked for some reason 

For the last problem, a line of symetry is when everything of either side of a line is the same, so if you folded the figure on the line everything would line up. Think about how you could split the octagon in half... maybe draw a line through the middles of two opositie sides? Or from opposite corners?