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 :)