NotThatSmart

The formula for the area of the parralleogram is \(base*height\), just like a rectangle! 

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Since we know AD bisects BAC, we have

BD  = 3 - x

CD =  x

Now, we can write

\(BD / AB = CD / AC \\ (3 - x) / 4 = x / 5 \\ 5 (3 - x) = 4x \\ 15 - 5x = 4x \\ 15 = 9x \\ x = 15/9 = 5/3 = CD\)

This just gives us

\([ ADC ] = (1/2) (CD) ( AB) = (1/2) ( 5/3) ( 4) = 10 / 3\)

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So we know that

\(CD = BD = 10 \\ ED = ED\)

Through this, we can find that triangle CDE is congruent to triangle BDE.

This means BE = CE!

\(sin CED / CD = sin CDE / CE \\ sin 75 / 10 = sin 90 / CE \\ CE = 10 sin 75 = BE ≈ 10.35\)

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The number of students ONLY in math class is a-b. 

The number of students ONLY in science class is c-(a-b) = c-a+b. 

The number of students ONLY in science class is just c-a. 

So, we just have a-b + c - a = c - b. 

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The freshmen are able to stand in 4! = 24 ways. There are 4 ways we can put them into the line with them all standing together. 

The sohphomores can staned in 3! = 6. 

We just have to do 24*4*6 = 576. 

There are 576 ways we could arrange them!

Thanks! :)

(b) For question b, we basically do the same exact thing as stated above. 

First, let's review the cases for \(a=b=c \). 

There are obviously n such triplets. 

Next, let's look at when exactly two are the same. You have 3 choices for which number is repeated and then n choices for the repeated number, and then n-1 for the distinct number. This gets us 3n(n-1). 

Lastly, we have 3 different numbers. There are \(6\)\(n \choose 3\) ways because there are 3 distinct numbers out of n and 3! = 6 ways to do this. 

Thus, when we add them together, we get \(n + 3n(n - 1) + 6 \binom{n}{3}.\)! 

Thanks! :)

First, let's see how many ways there are to arrange the 3 psychology books. 

We have 4p3 ways the arrange the books multiplied by the number of ways to put these books into the group of 7, which is 4. 

We have 24*4 = 96. 

Now, there are 4p4 ways to arrange the math books, which is simply 24. 

96*24 = 2304. 

There are 2304 ways to arrange the books! 

Thanks! :)

We would just have \(3!*4! = 6*24= 144\) ways, since we must place a children down before completing the problem. 

Thanks! :)

Alright, I'll try to tackle this problem!

(a) We can try to count the number of ordered pairs (a, b) ourselves. We can split these into two cases. 

1. \(a=b\)

2. \(a \neq b\)

1. Since we can choose any numbers between 1 and n, there are obviously n possibilities we can achieve. 

2. Since we can choose any 2 numbers between 1 and n that are not equal, we just have \(n \choose 2\). 

Hence, we just have \(n+\)\(n \choose 2\) \(+ \)\(n \choose 2\), which is exactly what we wanted in the first place. 

I hope I answered part (a) correctly!

Thanks! :)