MaxWong

(a) and (f)

\((2x-1)^5 = \displaystyle\sum_{r=0}^{5}\binom5 r\cdot2^rx^r\cdot(-1)^{5-r}\)

Term in x^2 = \(\displaystyle \binom 5 2 2^2\cdot x^2\cdot (-1)^{5-2}=-40x^2\)

Therefore the coefficient is -40.

This is trivial using binomial theorem (or simply using Pascal's triangle).

(p-q)^4 = p^4 - 4p^3*q + 6p^2*q^2 - 4pq^3 + q^4

The problem can be divided into:

(1) How to choose 4 children from 6 children;

(2) How to choose 2 adults from 4 adults.

Number of ways

\(= \displaystyle\binom6 4\cdot \binom4 2\)

\(=\dfrac{6!}{4!(6-4)!}\cdot\dfrac{4!}{2!(4-2)!}\\ =\dfrac{6!}{(2!)^3}\\ =90\)

\(5b^2+5b+4 = (2b+4)^2\\ 5b^2+5b+4=4b^2+16b+16\\ b^2-11b-12=0\\ (b-12)(b+1)=0\\ b=12 \text{ or } b=-1\text{(rejected)}\\ \therefore b = 12\)

Your hypothesis is correct. One must be 0 for the mode to be 0.

So now the numbers are 0 0 0 2 2 6 7 7 9.

If one more number is added, the number of integers in the list will be 10.

Let x be the other integer.

The numbers are 0 0 0 2 2 6 7 7 9 x.(I don't know the value of x, without loss of generality, I put it at the back of the list.)

Now the mode apparently is not 3.5, Therefore x should be added right after the left half of the list. (why?)

So the numbers now are 0 0 0 2 2 x 6 7 7 9.

Next thing we do is to solve \(\dfrac{2+x}{2} = 3.5\)

Just by inspection, x = 5.

Therefore the 2 numbers added are 0 and 5.

\(\boxed{\text{Note: } \\\text{speed} = \dfrac{\text{distance}}{\text{time}}\\\text{distance} = \text{speed}\cdot\text{time}\\\text{time}=\dfrac{\text{distance}}{\text{speed}}}\)

\(\text{Let }d\text{ miles be the distance between Washington DC and Baltimore.}\\\quad\;\;\;\!\;\! t\text{ hours be the time of return trip.}\) 

(magic: we can solve that without knowing the actual distance and time)

\(t = \dfrac{d}{50}\)

Time taken by Washington DC -> Baltimore = d/60

Total time used = \(d(\dfrac{1}{60}+\dfrac{1}{50})\)

\(\quad\text{Average speed} \\= \dfrac{d}{d\left(\dfrac{1}{60}+\dfrac{1}{50}\right)}\\=\dfrac{1}{\dfrac{1}{60}+\dfrac{1}{50}}\\=\dfrac{300}{11}\text{mph}\)

Firstly, the word "hypotenuse" is defined by "the longest side of a right triangle". Therefore the word hypotenuse implies that the triangle is a right triangle.

However, the condition is not enough to solve the problem. The shortest side can be infinitesimally close to 0 and the other leg of the right triangle can be infinitesimally close to 15 without another bound.

I assume the problem is asking about the smallest number of a Pythagorean triple with greatest number "15". If so, the answer will be (9,12,15), where the length of shortest side is 9.

When x = 0, y=f(x) is not connected because of a simple hole that could be plugged by adding or moving one point.

When x = 1, y=f(x) is connected and smooth.

When x = 2, y=f(x) is connected and with a corner.

When x = 3, y=f(x) is not connected because of a simple hole that could be plugged by adding or moving one point.

When x = 4, y=f(x) is not connected because of a vertical jump that could not be plugged by moving one point.

When x = 5, y=f(x) is connected and smooth.

When x = 6, y=f(x) is not connected because of a simple hole that could be plugged by adding or moving one point.

The probability of (\(4\leq x\leq 8\)) is just the area under the graph between x = 4 and x = 8.

P(\(4\leq x\leq 8\)) = (3+4)*(0.1)/2 + 4*0.05 = 0.35 + 0.2 = 0.55.