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We calculate the trailing zeroes of each of 100!, 200!, 300!, and 400!.

Note that we have this nice formula for calculating the trailing zeroes of n! : \(\#\text{ trailing zeros} = \displaystyle\sum_{k = 1}^\infty \left\lfloor\dfrac{n}{5^k}\right\rfloor\).

Then \(100!\) has \(\left\lfloor\dfrac{100}5\right\rfloor + \left\lfloor\dfrac{100}{5^2}\right\rfloor = 24\) trailing zeros.

Similarly, 200! has 49 trailing zeros, 300! has 74 trailing zeros, and 400! has 99 trailing zeros.

Then, (100!)(200!)(300!)(400!) has 24 + 49 + 74 + 99 = 246 trailing zeros.

We have: \(32x^3 = 2\)

Dividing both sides by 32, we get: \(x^3={1 \over 16}\)

Taking the cube root of both sides, we have: \(x = {1 \over { \sqrt[3]{16}}} = \color{brown}\boxed{2 \sqrt[3]{2}}\)

Simplifying the left hand side gives 32x^3 = 2.

DIviding by 32 on both sides and taking cube root gives the answer.

Answer: \(x = \dfrac1{2\sqrt[3]2}\)

Hint: If you project the vector \(\overrightarrow{AV}\) onto the line, take the projection as \(\overrightarrow{p}\), \(\overrightarrow{p}\) is actually \(\overrightarrow{OP} - \overrightarrow{OA}\).

\(4^x - 2^x = 56 + 11\cdot 2^x + 2^{x - 1}\)

If we let \(t = 2^x\) we have 

\(t^2 - t = 56 + 11t + \dfrac12 t\\ 2t^2 - 25t - 112 = 0\\ (t - 16)(2t + 7) = 0\)

It is not possible that 2^x = -7/2, therefore 2^x = 16, which corresponds to x = 4.

Pick a pair of equations and solve the system of those equations, and you will get the vertices.

\(\begin{cases}y = -x-5\\x=2\end{cases} \implies (x, y) = (2, -7)\)

\(\begin{cases}y = -x-5\\y = -\dfrac15x + \dfrac{13}5\end{cases}\implies (x, y) = \left(-\dfrac{19}2, \dfrac{9}2\right)\)

\(\begin{cases}y = -\dfrac15 x + \dfrac{13}5\\x=2\end{cases} \implies (x, y) = \left(2, \dfrac{11}5\right)\)

We find the center of the circle by solving the system 

\(\begin{cases}(x - 2)^2 + (y + 7)^2 = \left(x + \dfrac{19}2\right)^2 + \left(y - \dfrac92\right)^2\\ \left(x + \dfrac{19}2\right)^2 + \left(y - \dfrac92\right)^2 = (x - 2)^2 + \left(y - \dfrac{11}5\right)^2\end{cases}\)

which just says the squared distance from the center to each vertex is the same.

Solving gives the center coordinates \(\left(-\dfrac{49}{10}, -\dfrac{12}5\right)\).

Radius of circle is just the distance between \(\left(-\dfrac{49}{10}, -\dfrac{12}5\right)\) and any one of the vertices, i.e., \(\sqrt{\left(-\dfrac{49}{10}-2\right)^2+\left(-\dfrac{12}5-\dfrac{11}5\right)^2} = \dfrac{23\sqrt{13}}{10}\).

We use center-radius form of circle equation, and we have the answer: \(\left(x + \dfrac{49}{10}\right)^2 + \left(y + \dfrac{12}5\right)^2 = \dfrac{6877}{100}\)

If you want a simpler answer, multiply both sides by 100 and get \((10x + 49)^2 + (10y + 24)^2 = 6877\).

Exercise: Fill in the missed details in my solution.

The exterior angles of the polygon form an arithmetic sequence. 

Then the interior angles of the polygon should also form an arithmetic sequence.

The first term of that arithmetic sequence is 136 and the last term is 145, and there are k terms.

Suppose that the common difference is d degrees. Then \(136 + (k - 1) d = 145\).

That means \((k - 1)d = 9\). Then k - 1 is a factor of 9. Therefore, k can only be 2, 4, or 10.

Obviously, 2-sided polygons are not a thing, and the interior angles are too large for a quadrilateral.

Then k = 10.

You can try to figure out how to present this in a rigorous way.

Where is a?

Anyways, when x approaches -3, the value of f(x) should approach -3 - 6 = -9 from the right.

If f(x) is continuous, then the value of f(x) should approach -9 from the left too.

Then 

\(\displaystyle\lim_{x\to -3} (3x + b) = -9\\ -9 + b = -9\\ b = 0\)

_ _ _ _ 0 -> 4! / 2 possibilities = 12 p.

_ _ _ _ 2 -> 12 p. - 3 p. (bc 0)  =  9 p.

_ _ _ _ 4 -> 12 p. - 3 p. (bc 0)  =  9 p.

                                                    ____

                                                    30 p.

So there are 30 ways.

\(\begin{cases}a^3 + 3ab^2 = 679\\3a^2 b + b^3 = -679\end{cases}\)

Adding gives 

\(a^3 + 3a^2 b + 3ab^2 + b^3 = 0\\ (a + b)^3 = 0\)

So a + b = 0.

Just expand it:

(2u^2 - v^3)^4 = (2u^2)^4 - 4(2u^2)^3 (v^3) + 6(2u^2)^2 (v^3)^2 - 4(2u^2) (v^3)^3 + (v^3)^4

The bold term is the term in u^2 v^9. You can just simplify it and you will see that the coefficient is -8.

Generally, to draw a vector (x, y), pick a starting point (a, b), draw an arrow from (a, b) to (a + x, b + y).

c is the vector representing a + b.

What is P?

We need to know a and n first before we can find c.

Note that \(\begin{cases}an = -20\\\dfrac{a^2 n(n - 1)}2 = 180\end{cases}\). (Use binomial theorem and compare coefficients to get this system of equations.)

Solving gives a = -2 and n = 10.

Now, note that \(c = \dfrac{a^3 n(n-1)(n-2)}{6}\). Substitute the values of a and n and simplify.

\(\dfrac{\sqrt{6y + 2}}{\sqrt{2y + 2}} = \dfrac52\\ \dfrac{6y + 2}{2y + 2} = \dfrac{25}4\\ \dfrac{3y + 1}{y + 1} = \dfrac{25}4\\ 4(3y + 1) = 25(y + 1)\\ 12y + 4 = 25y + 25\\ 13y = -21\\ y = -\dfrac{13}{21}\text{ (rej.)}\)

There are no solutions.

\((\frac{1}{8})^3 = \frac{1}{512} = 2^a \ |\cdot 512 \\ 1 = 2^{a+9}\)

We can always express 1 as n² ! Let n be 2:

\(2^0 = 2^{a+9}\)

because if n^{m + a} = n^{d + k}, then it applies m + a = d + k:

 0 = a + 9 | - 9

-9 = a

a = -9

That's the only possible solution.

Adding all 4 equations gives us: \(10(a+b+c+d)=0\)

Dividing both sides by 10, we find \(a+b+c+d=\color{brown}\boxed{0}\)

There are 3 cases to consider: 2 white balls, 3 white balls, and the Powerball

Powerball: There is a \(1 \over 10 \) chance of this happening, and it is a win automatically

2 white balls: There are \({10 \choose 3 } = 120\) ways to choose the 3 numbers. There are \({3 \choose 2} = 3\) ways to choose the 2 numbers that will be drawn. We also need to pick one of the 7 losing numbers. Thus, the probability of this happening is \({ 7 \times 3 \over 120} = {7 \over 40}\)

3 white balls: There are \(10 \choose 3\) ways to choose the 3 numbers. There is \({3 \choose 3 }= 1\) way to choose the 3 numbers. Thus, the probability of this is \(1 \over 120\)

Thus, the total probability is \({1 \over 10} + {7 \over 40} + { 1 \over 120} = \color{brown}\boxed{17 \over 60}\)

Thanks!

How'd you get that answer?

The possible values of |a + b + c| are 1, sqrt(3), and 3.

This is incorrect. Two sides have to be larger than the other side (triangle).

a + b > c

a + c > b

b + c > a

If that does not apply, it is not constructible.

34,650 a^8 b^8 c^8

Use the 7th row of 'Pascal's triangle'.

Expand and get:

64x⁶ - 192x⁵ + 240x⁴ - 160x³ + 60x² - 12x + 1