justchillin

\[\sin^2 θ = 1 - \cos^2 θ = 1 - \frac{4}{13} = \frac{9}{13}\]

every angle in a parallelogram is between 0 and π radians, so θ must be positive, so

\[\sin θ = \sqrt{\frac{9}{13}} = \frac{3}{\sqrt{13}}\]

area of parallelogram is

\[\|{a}\| \|{b}\| \sin \theta = \sqrt{65} \cdot \sqrt{5} \cdot \frac{3}{\sqrt{13}} = \boxed{15}\]

We can mutiply out (a-bi)² to get a² - 2abi - b².

Since a and b are positive integers, the -10i on the right side of the equation must come from -2abi.

Therefore -10i = -2abi so ab = 5.

And now we see this problem is literally impossible because 5 can only factor into 1 and 5 or 5 and 1, and (1 - 5i)² is -24 - 10i while (5 - i)² is 24 - 10i. So unless I'm dumb and missed something, which wouldn't suprise me, this question is either has a typo or is impossible. 

Assuming it's a typo and the left side is actually adds to 24 - 10i and not 21 - 10i, the answer is 5 - i.

Factoring x² + 7x + 12 ≤ 0, we get (x + 3)(x + 4) ≤ 0, so -4 ≤ x ≤ -3. Probaby not the right way to do it, but I then plugged in -4 and -3 to see which resulted in a higher value, and -4 gets us 16 - 20 + 6 = 2, and -3 gets us 9 - 15 + 6 = 0, so the answer is 2.

If the number is x, we can write \(\frac{2x}{x+3}=9 \Rightarrow 2x=9x+27 \Rightarrow 7x=-27 \Rightarrow x=-\frac{27}{7}\).

The x coordiante of the midpoint would be \(\frac{2-(-6)}{2}=\frac{8}{2}=4\) and the y coordinate would be \(\frac{5-3}{2}=\frac{2}{2}=1\) so the midpoint is (4,1) and the product of the coordinates would be 4 * 1 = 4

The prime factoriztion of 112 is 2^4*7, so we can factor out 2^3 and get \(\sqrt[3]{112}=2\sqrt[3]{\frac{112}{2^3}}=2\sqrt[3]{14}\). So the answer would be 2 + 14 = 16.

(9x^5+25x^3- 4) - (x^5 - 8x^3 - 4) = (9-1)x^5 + (25+8)x^3 + (-4+4) = 8x^5 + 33x^3

The circumcenter is (-2,1).