RiemannIntegralzzz

To finish with the same amount of wins as losses, we need to have $50%$ won and $50%$ lost.

Let's break up this season into 20% chunks. Let's denote the percent of games won in each chunk $c_i$, where $i$ denotes which chunk we are on, 1 through 5.

We know have $\frac{c_1+c_2+c_3+c_4}{4}=48$, and we want to have $\frac{c_1+c_2+c_3+c_4+c_5}{5}=50$

Multiplying the first equation by four, we get $c_1+c_2+c_3+c_4=192$.

Multiplying the second equation by five, we get $c_1+c_2+c_3+c_4+c_5=250$

Substituting in $c_1+c_2+c_3+c_4=192$, we get $192+c_5=250$, so ------->$c_5=58%$<-------(for some reason my \boxed{} isn't rendering)

*Note: All the c_i's were in percent form, so, say, $c_2=28$ means that we won $28%$ of our games in chunk $2$, $20%$ to $40%$ of the season.

Writing these complex numbers as points, we have $(-11, 3)$ and $(3, -7)$. The midpoint is given by the average of the x- and y-coordinates, so we have $(\frac{(-11)+(3)}{2}, \frac{(3)+(-7)}{2})$. Simplifying gives the point $(-4, -2)$. The complex number corresponding to that point is $\boxed{-4-2i}$

Just a side note, make sure to put $ around your LaTeX so it renders properly, and put \sqrt{} instead of sqrt().

Multiplying both sides by $\sqrt{r}$ gives $r-3=-2*\sqrt{r}*\sqrt{r}$. Simplifying the right hand side gives $r-3=-2r$, and subtracting r from both sides gives $-3=-3r$. Dividing by -3 gives $\boxed{r=1}$

Let the original photon with red light be $RP$, and the new photon with 8000 times the energy be $NP$.

Since $NP$ has $8000$ times the energy as $RP$, it has $\frac{1}{8000}$ times the wavelength of $RP$, because it is inversely proportional. The wavelength will then be $\frac{7*10^{-7}}{8*10^{3}}$

$\frac{7*10^{-7}}{8*10^{3}}=\frac{7}{8}*\frac{10^{-7}}{10^{3}}=\frac{7}{8}*10^{-10}=\frac{70}{8}*10^{-11}$(to get it in the format you want. Now,you have $a=\frac{70}{8}$ and $b=-11$, so $a+b=\boxed{-2.25}$

"Trigonometry is a branch of mathematics that studies relationships between side lengths and angles of triangles." - Wikipedia

The first things you will probably learn are sines, cosines, and tangents, which basically show how one angle relates to some side lengths using the functions sin(x), cos(x), and tan(x).

Here is an example(you will not need to understand, just a sneak peek of usefulness of trigonometry):

    Let us denote a right triangle with hypotenuse BC, and A as the vertex with the $90^\circ$ angle, B as the angle above the $90^\circ$ angle, and C as the rightmost angle. If $sin(B)=12$, and the hypotenuse is of length $2$, then what is side length $AC$?

    The answer is 24, because $sin(B)=\frac{AC}{BC}$

You do not need to understand this, this is just a sneak peek of the usefulness of trigonometry.

Hope this helps! For more definition, here is the link to the wikipedia article: https://en.wikipedia.org/wiki/Trigonometry

Taking the averages of the x and y coordinates gives the midpoint: $(\frac{2+(-6)}{2}, \frac{3+5}{2})=(-2, 4)$, so the product of the coordinates is $-2*4=\boxed{-8}$

Sorry, it was incorrect, but I have one more try.

Using the sum of cubes factorization on $a^3+b^3=45$ yields $(a+b)(a^2-ab+b^2)=45$, which gives $7(a^2-ab+b^2)=45$, which gives $a^2-ab+b^2=\frac{45}{7}$

Squaring $a+b=7$ gives $a^2+2ab+b^2=49$. We now have a system of equations! From here, there are two ways to proceed. One is to solve the system of equations, find the values of a and b, and plug them in. The other way, which is MUCH faster, will be shown here.

Subtracting the first equation from the second yields $2ab-(-ab)=\frac{298}{7}$, which gives $ab=\frac{298}{21}$

Now, we make use of the fact that $\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}$(you can check this yourself by splitting up the fractions), and we plug in! We now have $\frac{a+b}{ab}=\frac{7}{\frac{298}{21}}=\boxed{\frac{147}{298}}$

Sorry, but this was incorrect. Please show ALL YOUR WORK so I can check for errors. And also, yes, the problem asks to find x

Let my try posting it again...

Points $A$,$B$,$C$,$D$, are located on $\overline{AB}$ such that $AB=3AD=6BC$. If a point is selected at random on $\overline{AB}$, what is the probability that it is between $C$ and $D$? Express your answer as a common fraction.

Better?