NotThatSmart

We just have

\(64 \cdot 3,816 = 244,224 \)

\(244224 \pmod{ 64} \equiv 0\)

244224 works. 

Thanks! :)

First, let's move all terms to one side and combine all like terms. We get the equation

\(x^{2}+10x+9=0\)

Now, notice that we can easily factor the quadratic equation. We get

\((x+9)(x+1)=0\)

This gives us the values for x of 

\(x=-1\\ x=-9\)

Thus, our answer is -1, -9.

Thanks! :)

\(f(x) = \sqrt{x - \sqrt{x}}.\)

This is impossible. No such 3 digit numbers exist for x.

Let's make some observations of the problem. 

First, note that triangle ABC is a 45-45-90 right triangleThis means that\( AB = BC  =   12/\sqrt 2 =  6\sqrt 2 \)

Also, we have that

\(AD = AD\\ BD = BD \\ AB = BC \)

This means that triangles  ABD  and CBD  are congruent

Thus, we have \([ ABD ]   =  (1/2) [ABC]   =  (1/2) ( 1/2) ( 6\sqrt 2)^2  =  (1/4) (72)   = 18\)

So our aswer is 18. 

Thanks! :)

It's not as hard as it might seem!

We just simply have

\([ b^2 - a^2 ] / [ b - a ] = [ (b -a) (b + a) ] / (b -a) = b + a = 2\)

So 2 is the answer. 

THanks! :)

Let's use trig. 

\(slope PQ = -1 / tan [ (arctan (4) + arctan (5) ) / 2 ] = [19 - \sqrt {442} ] / 9 ≈ -0.225 \)

Thus, our answer is -0.225. 

Thanks! :)

Let's first graph this problem to have a visual on what we must do. 

First off, let's note we have a lot of right triangles to work with. This also means we almost certainly must use the pythaogrean thereom. 

So let's do it. First, we have the equation 

\((r -6)^2 + (r - 6)^2 = r^2 \\ 2r^2 - 24r + 72 = r^2 \\ r^2 - 24r + 72 = 0\)

Now, let's complete the square for r. 

\(r^2 - 24r = -72 \\ r^2 -24r + 144 = -72 + 144 \\ (r- 12)^2 = 72\)

since the distance cannot be negative, let's take the positive root for r. We have

\(r -12 = \sqrt{72} \\ r = \sqrt {72} + 12 \\ r = 6\sqrt 2 + 12 ≈ 20.49\)

So 20.49 is our answer. 

Thanks! :)

Let's start with what we should do for any geometry question, which is to of course, graph it. 

Note that we have right triangles ADC, ABE, ABD, BHD, BEC and triangles ABE and ADC are a 30-60-90 and a 45-45-90 respectively. 

Now, it's pretty simple from here. 

We first draw the altitude for triangle ABC from point C to line segment AB, it intersects point H along the way.

Also, since we know BAC is 60 degrees, HCA would be 30 degrees!

30 degrees!

Thanks!