mworkhard222

#!: Split the hexagon into 16 parts. 10 are shaded. 10/16 = 5/8. 5 + 8 = 13

#2: Split the hexagon like this: 

So each of the triangles is 60/8 = 7.5,  and 4 * 7.5 = 30. 60 + 30 = 90

using "x" this should be

x < 5

so as an interval: $\left(-\infty \:,\:5\right)$

it's a system of equations:

$\frac{1}{p}+\frac{1}{q}=1$ and $pq=9$

$p=\frac{9+3\sqrt{5}}{2}$

$q=\frac{9-3\sqrt{5}}{2}$

the sum is 9

umm??? still the second line is not working. it says:

(1/3)t < t + 3

idk why the first 2 lines are not showing up, it's supposed to say:

$\frac{1}{3}\cdot \:t-5<\:t-2$

\(\frac{1}{3}t

\(\frac{1}{3}\cdot t-5

\(\frac{1}{3}t

$-\frac{2}{3}t<3$

$\frac{2}{3}t>-3$

$2t>-9$

$t>-\frac{9}{2}$

So in interval notation: $\left(-\frac{9}{2},\:\infty \:\right)$

Well if the 2 angles are 30 and 60 degrees, this has to be a 30-60-90 right triangle.

Heres a pic i made:

This is quite complicated but if u solve this system of equations i think you get: $x=\frac{14}{3},\:z=\frac{8}{3},\:y=\frac{14}{3},\:L=\frac{26}{3}$

$\frac{14}{3}+\frac{8}{3}=\frac{22}{3}$

The slope intercept form is: $y=mx+b$

First you need to find the slope (m). $m=\frac{y_2−y_1}{x_2-x_1}.$

$m=\frac{−2−\left(−5\right)}{6−\left(−3\right)}$

$m=\frac13$

Next, you find the y-intercept, or b. You can plug in any of the 2 points like this: $y=\frac13x+b$

$-2=\frac13\cdot6+b$

$2+b=-2$

$b=-4$

So our final answer is $y=\frac13x-4$

(I just wanted to make ElectricPavlov's solution more clear)

HINT: There are 4 balls and 3 boxes, where all the balls are different. One of the 3 boxes has to have 2 balls.