gb1falcon

Since 2z and z^(3/2) cannot be combined to create a constant term in the given equation, the answer is 1 (1^4).

\(-8+6m-\frac{1}{2} ( -4m+16)\)

\(6m-8-(8-2m)\)

\(6m-8-8+2m\)

\(8m-16\)

That is the most it can be simplified.

Pretty good! Here's another similar way.

\(\sqrt(2)*\sqrt(6)*\sqrt(110)*\sqrt(120)*\sqrt(450)*\sqrt(520)\)

\(\sqrt(2*6*110*120*450*520)\)

\(\sqrt(37065600000)\)

\(100\sqrt(3706560)\)

\(800\sqrt(57915)\)

\(7200\sqrt(715)\)

\(7200*\sqrt(5)*\sqrt(11)*\sqrt(13)\)

What do you need help with figuring out?

16x-57=10x+3x

16x-57=13x

3x-57=0

3x=57

x=19

Then, 

16x-57=10x+3x=247

I don't know which answer you wanted, so I just gave both.

We can use complementary probability.

The cases do not matter, because the total probability of raining on Sunday is 20%, which has already been calculated for us.

The probability of it not raining on either day is 70% * 80% = 56%. So, the probability of it raining on at least one day is 100-56=44%.

As a sidenote, here is more information.

First, calculate the probability of it raining on Sunday. The probabilities of it raining on Sunday average 20%, so they add to 40%. One is twice the other, so if it rains on Saturday the probability is 80/3% and if it does not it is 40/3%.

I'm sorry, I forgot to include the side length of the square. I must have missed typing it. The side length of the square is 10. Does that change your answer?

C^m(x)=C(C(C(C ... (x))))..., performing C m times. Do this to 9:

9 is odd, C(9)=28.

28 is odd, so C(C(9))=14.

14 is even, so C(C(C(9)))=7. 

Continue to get 22,11,34,17,52,26,13,40,20,10,5,16,8,4,2,1

After applying C 19 times, we have reached 1. Therefore, the smallest m is 19.

I get about -0.99384, using a calculator.

What we need is to separate the 90% into some r (raincoats) and u (umbrellas)

2r+20=u+10

2r+10=u

Since u+r=90, we can add r to both sides.

3r+10=u+r

3r+10=90

3r=80

r=80/3

Then u = 190/3

If we add 30/3 to both u and r we get

u=220/3

r=110/3

So u is twice r.

Then, if 190/3% people brought umbrellas only, than 63.3333% of the people only brought umbrellas. You could also put 63 1/3%. Then, the number of students must be a multiple of 300 for this to work.