CitrusCornflakes

First of all, the two large rectangles at the side have sidelengths of 48 and 25. The total area of these are 2400cm^2. The front and back triangle have 2*1/2*14*24=336cm^2. Finally, the bottom face has area 14*48=672cm^2. 

We sum these together to get 3408cm^2.

:)

By your definition of the binary '*' operator, 3*-2 = 3^2 - (-2)^3 = 9 - (-8) = 9 + 8 =17

Your answer is 17.

Draw a square with sidelength a.

Draw a square with sidelength b such that b

Area of one triangular face =bh/2 = 6*4/2=12cm^2.

Area of triangular faces = 24cm^2.

Area of one rectangular face = bh = 40cm^2.

Area of rectangular faces = 120cm^2.

Add them together. 144cm^2.

For the compound interest question, notice how it is compounded quarterly, which essentially means compounding by an extra $\frac{7}{8}$ percent per quarter. So just raise $100+\frac{7}{8}$ as a percent to the power of 20 (20 quarters becomes 5 years) and multiply that by the starting \$2500. So use the equation $2500*(1+0.035/4)^20$ to evaluate. That gives you an approximate of $2975.85$ dollars.

stonks go brrrr

Please post only one question per thread, but here goes.

For the trig question, then you notice that the hypotenuse of the right triangle is 200, and the height is 40. We then see the angle of elevation is just the sine of 40/200, or simplified: $sin(\frac{1}{5})$. 

4f(2x)=(8,8)

So then f(2x)=(2,8) by height diminishing.

Then, by length dimishing, f(x)=(2,4).

The sum of coordinates is 6

gimme them upvotes

Clearly, &4=9. Then, #9=81. Sp #(&4)=81.

:)

Hello all, just a few resources for using LaTeX.

I found a great resource on AoPS about LaTeX. It's not a gigantic read, so I suggest that if you want to learn LaTeX to a deeper extent to look at this webpage. It also covers aligning equations.

https://artofproblemsolving.com/wiki/index.php?title=LaTeX:LaTeX_on_AoPS

If you want to create LaTeX Tables, use this webpage to generate one.

https://www.tablesgenerator.com/#

The only problem is that webpage doesn't allow math input for the tables.

You can also learn to make them yourself in the below link.

https://www.overleaf.com/learn/latex/tables

If you don't know how to do some LaTeX symbols, that's no problem. You can just directly search them up online, and there will definitely be some great sources.

If you are making complicated, hard-to-read equations in LaTeX, then don't spend your time getting confused over the syntax! All you got to do is to type the equation up in the Desmos Graphing Calculator! Copy the result, and then if you paste it, then it will be in LaTeX. All you have to do is to wrap it with dollar signs.

https://www.desmos.com/calculator

-CitrusCornflakes :) $\int_{0}^{\pi}\sin\left(x\right)\ dx\ =\left(-\cos\left(\pi\right)+C\right)-\left(-\cos\left(0\right)+C\right)=2$

A hint:

The number of trailing 0s is the number of factors of 5 in the whole expression. So if you can work out an approach to find out all the factors of 5, then you've solved it.

-CitrusCornflakes :)

We know all factorials 10 and afterwards end in 00 (as there are two factors of 5 and enough factors of 2s), so it is really 5! which counts. This ends in 20, so the last two digits of the described following sum is clearly 20.

Answer:20

-CitrusCornflakes :)

@above, simply functional evaluation. For p(1,-1), then the case is 'otherwise' so we have 3*1-1=2. For p(-5,-2) we have case 2, so then -5-2(-2) equals -1. The outer function then has arguments 2 and -1, which is in the case 'otherwise' so we have 6-1=5.

So the final functional output is 5.

So we have side CD and side BD. We want to find side BC first to work out tangent. Just use pythagoream theorem, so BC=$\sqrt{5}$. Now, you use that Tangent=Opposite/Adjacent so you have CD/BC which is $\frac{\sqrt{14}}{\sqrt{5}}$.

:)

@cating

The problems from AoPS Mainly come from special classes, where the problems are not allowed to be shared. I don't think there is another way to access other problems either...

Probably the only thing we can actually do is to just watch out for problems during competition times.

We notice the value under the square root must be positive to be in the domain. So we have $(x+3)^2-(x-8)^2\geq0$. We see that after simplifying, $22x-55\geq0$, where the minimum of $x$ is then $2.5$. Clearly, the smallest $x$ in the domain is $2.5$.

When you state that adjacent cubes have different colors, then the layers must be like this:

First layer:

Second Layer

If we inverse the colours, then we can rotate the new cube 90 degrees to get the original cube we first drew.

So there is only $\boxed{1}$ possible cube.