BuilderBoi

That was my answer

https://web2.0calc.com/questions/counting-and-probability_7

First, simplify the equation to \(x^2 - 3x + 4 = 0\)

Note that \(a^3b + b^3 a = ab(a^2 + b^2) = ab((a+b)^2 - 2ab)\)

By Vieta's, \(a + b = {b \over a} = -{ -3 \over 1} = 3\) and \(ab = {c \over a} = 4\)

Can you take it from here?

Here's my attempt: 

Because \(AB \parallel CD\), \(\angle DAB = \angle ADC = 50^\circ\).

Now, draw line segment \(CB\), and label the intersection point \(E\).

Note that \(\triangle ECD \) is isosceles, so \(\angle ECD = 50^ \circ\). This means that \(\angle ECD = \angle AEB = 80^ \circ\)

So, the length of \(\overset{\large\frown}{AB} = {80 \over 360} \times 2 \times 18 \times \pi = \color{brown}\boxed{8 \pi }\)

I'll do the numerator, and you can do the denominator. 

For the numerator, let \(x = 0.{ \overline{72}}\). This means \(100x = {72. \overline{72}}\) (move the decimal point 2 places to the right)

Subtracting the 2 yields us \(99x = 72\) (the repeating parts cancel out)

Dividing by 99, we find that \(x = {72 \over 99}\)

Your answer is wrong, the number is more than the denominator, so the fraction will be greater than 1...

The expected value is \(({3 \over 4} \times 1) + (-1 \times {1 \over 4}) = \color{brown}\boxed{1 \over 2}\)

https://web2.0calc.com/questions/need-help-builderboi

By trial and error, x = 3 and y = 1 

Combine like terms: \({1 - i \over 8i -2}\)

Multiply by \({8i + 2 \over 8i + 2}\): \({(1 -i)(8i + 2) \over {(8i - 2)(8i + 2)}}\)

Expand: \({-8i^2 + 6i + 2 \over 64i^2 - 4}\)

Substitute \(i^2 = -1\): \({8 + 6i + 2 \over -64 - 4}\)

Combine like terms: \({10 + 6i \over -68}\)

Simplify: \(\color{brown}\boxed{- 5 - 3i \over 34}\)