power27

By the pathogorean theorum, the hypotonuse of the first triangle is \(\sqrt{7^2+24^2} = 25\)

Because the two triangles are similar, the rario between the side lengths will be the same on both triangles.

\(\frac{100}{short leg}=\frac{25}{7}\) 

cross multiply

700=25*short leg

divide by 25

short leg of the second triangle=\(\boxed{28}\)

There are 4!=24 ways to draw the slips out of the hat.

There are only 2 ways to draw C first and D last:

CABD

CBAD

because once you decide when you draw C and D, the only two papers you have to worry about ordering are A and B in the middle and there are only 2 ways to do that.

So the probability is \(\frac{2}{24}=\boxed{\frac{1}{12}}\)

To have exactly one factor that is prime, the number must be prime so the prime factorization is the number*1, because 1 is not prime.

The divisibility rule for 11 is to take the alternating sum and difference of the digits and if that is divisible by 11 so is the number.

Take the 4 digit palendrome ABBA and using this rule on it you get A-B+B-A=0, and 0 is divisible by 11 so every four digit palendrome is divisible by 11. No 4 digit palendromes have exactly one digit that is prime.

The only way to interperate this problm so that it has an answer is if you can repeat factors.

11*11*11 = 1331 is a four digit palendrome, and it also as only one unique prime factor.

Are the books distinguishable?

All these questions look like they are from some type of online test...

First write out all the cases:

Odd: 3x-1

Even(not divisible by 4): .5x+1

Divisible by 4: .25x

Dorthy could have entered any type of number, so set 13 equal to what the button could do to the number she entered:

13=3x-1

14=3x

14 is not divisible by 3, so she did not enter an odd number.

13=.5x+1

12=.5x

x=24

24 is divisible by 4 so she could not have entered a number only divisible by 2.

13=.25x

x=52 so \(\boxed{52}\) is the only number she could have plugged into this function.

There are 9 choose 4 = 126 ways to choose the first team.

For the next team, there are 5 remaining people to choose from so there are 5 choose 3 = 10 ways to make it.

The remaining 2 people will be the 2 person team.

126*10= \(\boxed{1260}\) ways to choose the teams.

The top right corner is Mack's goal. Bottom left is where he starts and the red spot is the spider. Blue numbers are the number of ways Mack could get to the point.

He starts at (0,0) and there is only one way to get there. There is also only one way for him to get to all the points on the bottom and left edges, because he can only move up and right.

To get to (1,1), there are 2 ways to get there because he could come from (0,1) or (1,0).

To get to (2,1) he could come from (2,0) or (1,1). The number of ways he could get to (2,1) is the sum of the ways he could get there from the point below or the point to the left so 2+1=3 ways to get to (2,1).

If you look at the drawing, all the blue numbers are the sum of the blue numbers 1 below and 1 to the left. You might notice that this looks like Pascal's triangle! (Because it pretty much is).

Continuing to sum the numbers in this manner, the number of ways to get to (5,7) is \(\boxed{426}\)

There are other ways to solve this problem using combinations, and if this is a homework problem that is probabally the way you know how to solve it. I like this way better .

The last digit of a multiplication problem is the last digit of the product of the two unit digits.

First make a list of the unit digits of the sequence and see if there is a pattern.

1, [2, 4, 8, 2, 6, 2], [2, 4, 8, 2, 6, 2], [2, 4, 8, 2, 6, 2]

It looks like after 1 there is a 6 digit repeating pattern [2, 4, 8, 2, 6, 2]

Ignoring the 1 at the beginning of the sequence the problem is now find the 34th term in the pattern that was found.

The largest multiple of 6 under 34 is 30 so the 30th term will be the last 2 in the pattern.

30- 2

31- 2

32- 4

33- 8

34- 2

The unit digit of the 35th term in the original sequence is \(\boxed{2}\)