proyaop

First we have to square root both sides, thus, $z = \pm\sqrt{2i + 2}$.

$\sqrt{2i + 2} = (2i + 2)^{1\over2}$

Then, opening paranthesis is $\sqrt{2i} + \sqrt{2} + 2i$.

That simplifies to $\sqrt{2} + i\sqrt{2} + \sqrt{2} + 2i$ and $-\sqrt{2} - i\sqrt{2} + \sqrt{2} + 2i$

Thus, the complex numbers of z are $2i + 2\sqrt{2} +i\sqrt{2}$ and $2i - i\sqrt{2}$.

or maybe im imagining things :)

The segment across from point A, where point D lies on is the shortest side, and is side BC. Thus, if side BC is 10, the other two sides are 15 and 30.

But I think there might be a typo, because than this triangle would be degenerate using the triangle inequalities...

Guest please repost your question, because a triangle CAN'T have side ratio of 2:3:6. Please double check your post and make sure...

Thank you Guest! 

Yes you are very right.

For anyone reading this solution, x has to be a real number as declared in the question, thus x^2 - |x^2| could not be negative... 

x can't be i.

20^20 prime factorized is 2^40 x 5^20. Thus, using the prime factorization trick to find the # of factors, 20^20 has 861 factors. 

10^18 prime factorszed is 2^18 x 5^18. Thus, using the prime factorization trick to find the # of factors, 10^18 has 361 factors. 

Since all factors greater than 10^18 in 20^20 are a multiple of 10^18, thus there are 861 - 361 + 1 = 501 factors that are a multiple of 10^18 when choosing one from 20^20. 

(The reason I had a "+ 1" is because one factor of 10^18 is 10^18, which is also a multiple of 10^18. Thus there are 360 factors that do not work in 20^20.)

That means, the probability of choosing a divisor from 20^20 and having it be a multiple of 10^18 is 501/861.

Simplifying, we get the probability as 167/287.

You said it XxmathguyxX.

Remember Guest, the formula for the area of a trapezoid is the combined bases x the height x 0.5.

Great step-by-step process and well detailed explanations.

Great job BuilderBoi! Keep working hard!

We have ourselves here a trapezoid. The area of a trapezoid is h(base1 + base2) / 2.

Plugging in the values we see the area is 6(12 + 15)/2 = 3(12 + 15) = 3(27) = 81.

Thus, the area of the trapezoid is 81 units squared. 

I think the way to go for this problem is to substitute in a few values in terms of other variables. 

I multiplied equations 1 and 2 to get ${x^2y^5z\over{xz}}=720$.

Thus, $xy^5 = 720$.

Not quite sure what to do from here, maybe you can give it a go Guest! (try substituting in variables in terms of other variables, and try plugging it into the expression)

Good luck :D!

ABCD is a rectangle, so that means AB is one of its sides. AB has length 4. That means the other side of the rectangle has length 32/4 = 8 units. That means the x coordinate of point C and point D has to be 8 units away from the origin. 

Note: In the problem it says x > 0 so x cannot be -8.

Thus, x = 8.

Oh yes! Thank you for your input everyone! But I caught a quick error! The solution ITSELF is a + bi, the solutions AREN'T a, and b. Thus, we might have to use the quadratic formula to find the solutions.

Applying the quadratic formula $x = {-b \pm \sqrt{b^2-4ac} \over 2a}$. We get -5 +- sqrt(25 - 72) / 2. 

Thus the two solutions are $-{5\over2}+{\sqrt{-47}\over2}$ and $-{5\over2} - {\sqrt{-47}\over2}$. 

In this case, it doesn't matter whether b is negative or positive since we are squaring it eventually. a is 5/2. Thus, a^2 = 25/4 and b^2 = -47 or -188/4. 

That means, $a^2 + b^2 = -{163\over4}$, which approximates to -40.75.