chrissy

here's a hint: reflect the triangle around C so that AC is on BC and look for similar triangles

i'll write out a whole solution in a bit but that's mainly what you need to figure it out

we note that \((y+2)(2y+1) = 2y^2 + 5y +2\), so we plug that in and solve for y:

\(\begin{align*} \frac{1}{y+2}-\frac{4}{2y+1} + \frac{3y-4}{2y^2 + 5y + 2}&= 0 \\ \frac{2y+1}{2y^2 + 5y + 2} - \frac{4(y+2)}{2y^2 + 5y + 2} + \frac{3y-4}{2y^2 + 5y + 2}&= 0 \\ \frac{2y+1-4y-8+3y-4}{2y^2 + 5y + 2}&=0 \\ \frac{y-11}{2y^2 + 5y + 2}&=0 \\ y-11&=0 \\ y&=\boxed{11} \end{align*} \)

note: since we multiply by \(2y^2 + 5y +2\), we have \(y \neq -2, -\frac{1}{2}\), but since \(y+2\) and \(2y+1\) are already denominators, we don't have to worry about fractions over zero.

if the points were to form a rectangle, the slopes would have to be perpindicular, or be negative reciprocals of eachother. we find the slopes of all the lines:

\(\begin{align*} \overline{AB} &\Rightarrow \frac{4-1}{0+3} = \frac{3}{3} =1 \\ \overline{BC} &\Rightarrow \frac{4-2}{0-0} = \text{vertical line} \\ \overline{CD} &\Rightarrow \frac{2+1}{0+3} = \frac{3}{3} =1 \\ \overline{DA} &\Rightarrow \frac{-1-1}{-3+3} = \text{vertical line} \\ \end{align*}\)

in order for the lines to be perpindicular, the slopes of \(\overline{AB}\) and \(\overline{CD}\) would have to be 0, which would make a straight line, instead of 1, so the points do not make a rectangle. however, we have two pairs of parallel sides, so we know we have a parallelogram.

cylinder A has a volume of \(\left(\frac{d}{2}\right)^2\pi h = 14^2\pi \cdot 15 = 2940\pi\) and cylinder B has a volume of \(\left(\frac{d}{2}\right)^2\pi h = 11^2\pi \cdot 159= 2299\pi\). this means that cylinder B is \(\frac{2299}{2940} \cdot 100 \approx 0.7819 \cdot 100 \approx 78.2\%\) percent of cylinder A, which means cylinder A will end up being 

\(100-78.2=\boxed{21.8\%}\)

percent full after the pumping.

for a number to be divisible by 25, it must end in either 00 (making it a multiple of 100), 25 (a multiple of 100 + 25), 50 (a multiple of 50), or 75 (a multiple of 100 + 75). since 0 isn't in our digits, the digits must be in the form

\(\times\times\times\times25\)

or

\(\times\times\times\times75\)

where \(\times\) represents any digit not used. there are \(4!\) ways to arrange the digits that fit the first form and \(4!\) ways for the second, so we have a total of 

\(4! + 4! = 24 + 24 =\boxed{48}\)

ways to arrange the digits.

can you edit/comment on the posts with the lengths and variable names? i think they got fussed up in the conversion of \(\LaTeX\).

the number of trailing zeros is determined by how many powers of \(10\) are in the final product. \(25\) has two powers of \(5\), \(150\) has two powers of \(5\) and one power of \(2\), and \(2008\) has three powers of \(2 \). once we count the exponents, we have a total of nine powers of \(2\) and eighteen powers of \(5\), which gives us a total of nine powers of \(10\), giving us \(\boxed{9}\) trailing zeros.

if one of the angles is \(60^\circ\), then we know the rhombus is made up of four 30-60-90 right triangles, with the hypotenuses making up the perimeter. we know that the hypotenuse is \(8\div4=2\), so one leg has a length of \(1\) and the other leg is the length \(\sqrt3\). this means that the lengths of the diagonals are \(2\) and \(2\sqrt3\), so the area of the rhombus is \(2 \times 2\sqrt3 = \boxed{4\sqrt3}\).

since it's pairs, the only unit fraction that is large enough to fit this requirement is \(\frac12\), so the only pair is \((\frac12,\frac12)\).

AE = 70

just a bunch of similar triangles :)

\(\begin{align*} \sqrt{3} \times 3^{\frac{1}{2}} + 12 \div 3 \times 2 - 4^{\frac{3}{2}} &= \sqrt{3} \times \sqrt{3} + 8 - (\sqrt{2})^3 \\ &= 3 + 8 - 8 \\ &= \boxed{3} \end{align*}\)

we find the slope of the line:

\(\begin{align*} 4x+3y&=7\\ 3y&=-4x+7\\ y&=-\frac{4}{3}x+7 \end{align*}\)

and plug in the \(x\) and \(y\) of \((2, -11)\) to find the \(y \)-intercept:

\(\begin{align*} y &= -\frac{4}{3}x + b \\ -11 &= -\frac{4}{3}(2)+b \\ -11 &= -\frac{8}{3} + b \\ b &= -\frac{25}{3} \end{align*}\)

and plug in the slope for 

\(\boxed{y=-\frac{4}{3}x-\frac{25}{3}}.\)

\(|a+bi|^2 =( \overline{a+bi}) \cdot (a+bi)\), and since \(|w| = 1\), we know \(|w|^2 = |w| \cdot |w| = 1 \cdot 1 = 1 =|w|\), giving us 

\(\begin{align*} |w| &= 1 \\ |w| &= |w|^2 \\ |w| &= \overline{w}w \\ \overline{w}w &= 1 \\ \overline{w} &= \frac1w. \end{align*}\)

we can get \(\overline{z} = \frac1z\) through the same process. we make \(x = \frac{w+z}{1+wz}\), which means \(\overline{x} = \frac{\overline{w}+\overline{z}}{1+\overline{w}\cdot\overline{z}}. \)we plug in our values for the congujates of \(w\) and \(z\), which gives us

\(\begin{align*} \overline{x} &= \frac{\frac{1}{w} + \frac{1}{z}}{1+ \frac{1}{w} \cdot \frac{1}{z}} \\ &= \frac{\frac{z}{wz} + \frac{w}{wz}}{\frac{wz}{wz}+\frac{1}{wz}} \\ &=\frac{\frac{w+z}{wz}}{\frac{1+wz}{wz}} \\ &= \frac{w+z}{1+wz} \\ &= x \end{align*}\)

we get \(\overline{x} = x\), which is only true when \(x\) is real.

i believe the answer is 0 for all of them, since the only amount of spaces that you can move is even, and an even number + even number = even number and even number - even number = even number

edit: aaaaaah. misunderstood the problem. CPhill got it right!

supplementary angles = sum of angles is 180
complementary angles = sum of angles is 90
 

180-110=70 degrees :)