BuilderBoi

Multiply both the numerator and the denominator by \(y \over \sqrt3\), the reciprocal of \(\sqrt3 \over y\). 

This gives: \(xy\over \sqrt3\), which can be written as \(xy \sqrt3 \over 3\)

8.5/0.5 = 17

17 * 25 = ___

Melody, your diagram has a mistake...

\(\overline{AB} = 44\), not  \(\overline{CB}\)

Alright, I'll take a shot...

Reflect \(\triangle AME\) over angle bisector \(\overline{AD}\).

You now have 2 triangles, and the sum of the areas equals \(\triangle ACE\)

The first triangle has a height of 10, and a base of 22 (Note: It has the same side lengths as \(\triangle AME\) because it is a reflection)

The second triangle has a height of 10, and a base of 8 (both bases sum to 30). 

Thus, the area is \(110 + 40 = \color {brown} \boxed {150}\)

Here is the messy diagram I made: 

Write as a system of 3 equations: 

\(3a = b+c\)  (Equation 1)

\(5b = a+c\)  (Equation 2)

\(c=80+b\)  (Equation 3)

Sub in (Equation 3) into (Equation 2):  \(5b = a+b+80\)

Simplify:  \(b = {a \over 4}+20\)

Now, sub what we found for b, and (Equation 3) into (Equation 1): \(3a= {a \over 4} +20+80+ {a \over 4} +20\)

Solving, we find \(a = 48\)

Subbing the value of \(a\) we just found and (Equation 3) into (Equation 2), we get: \(5b = 48+80+b\)

Solving, we find \(b = 32\)

Plugging in what we found for the value of \(b\) into (Equation 3), we find that \(c = 112\).

Now do \(a + b+c\) to find your answer. 

You can write this as an equation, where n is the number of weeks. 

\({2n \over 12+3n} = {1 \over 2}\)

Solve for n, and round up to get your answer. 

The question is incomplete...

Anything that is equal to, or bigger than 65.

Let X be a square with side length 3 and A be 4. Scaling by a, we have a 12 x 12 square now. The area of this square is 12 x 12 = 144, \(4^2\) times bigger than the original square's area with 9. Thus, scaling by a increases the area by \(a^2\). (Note: If you still don't believe me, try this out with a different shape and different values for X and A.