NotThatSmart

Well, this problem can be solved by the pythagorean thereom. 

First off, note that AK is the altitude. This gives us multiple right triangles to work with. 

We have

\(AK = \sqrt { AC^2 - CK^2 } = \sqrt { 8^2 - 3^2 } = \sqrt { 55 } \\ AB = \sqrt { AK^2 + BK^2 } = \sqrt { 55 + 4 } = \sqrt { 59 } \)

So the answer is sqrt59. 

"Thanks! :)

The length of $\overline{AB}$ is $4,$ and the circle has a radius of $5.$  Find the length $AB.$

ummm, the answer is 4?

Thanks! :)

mmm...not sure about my solution, but here it is. 

First, we have the modulo 

\(a \equiv 5 \pmod{7}\)

Since we must find a+1 mod 7, let's add 1 to both sides of our equation to get

\(a+1 \equiv 5+1 \pmod7\\ a+1 \equiv 6 \pmod7\)

We can test numbers. 

Let's test 5, 19, 26

\(6 \equiv 6\pmod7\\ 20 \equiv 6 \pmod 7\\ 27 \equiv 6 \pmod 7\)

So 6 is our answer. 

Thanks! :)

Let's note that 1/17 is a repeating decimal. 

\(1/17 = 0.\overline{ 0588235294117647}\)

There are 16 repeating digits in the decimal. 

Dividing 4000 by 16, we get

\(4000/16 = 250\)

Since it is divisble, then we can conclude that the last digit of the repeating decimal is the correct answer. 

Thus, 7 is the final answer. 

Thanks! :)

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

Let's use a handy trick to solve this problem. 

First off, let's set a variable. Let's set \(X=2.\overline{57}\)

For that value of X, then we have that

\(100 X = 257.\overline{57}\)

Now, we subtract x from 100x. Note that the repeated part of the decimal cancels out. Thus, we have

\(\eqalign{100 X &= &\hfill257.57...\cr X &= &\hfill2.57...\cr \hline 99X &= &255\cr}\)

We end up witht he equation

\(99 X = 255\)

Simplifying and isolating x, we get 

\(X = \frac{255}{99}\)

Thus, 

The answer is \(2.\overline{57} = 2 \frac{19}{33}\)

Thanks! :)

umm...not sure how to explain...the answer is

\(E21A800000_{16}\)

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

We have the inequality

\(4t^2-48+23t\le \:0\)

Now, let's complete the square for t. We get

\(4\left(t+\frac{23}{8}\right)^2-\frac{1297}{16}\le \:0\)

This format allows us to isolate t in a much simpler manner. Now, let's start the process. We get

\(4\left(t+\frac{23}{8}\right)^2\le \:\frac{1297}{16}\)

\(\left(t+\frac{23}{8}\right)^2\le \:\frac{1297}{64}\)

Now, square rooting both sides, it gives us two parameters to focus on for t. We get

\(-\sqrt{\frac{1297}{64}}\le \:t+\frac{23}{8}\le \:\sqrt{\frac{1297}{64}}\\ \frac{-\sqrt{1297}-23}{8}\le \:t\le \:\frac{\sqrt{1297}-23}{8}\)

Now, the final step is to write this in interval notation. We finally get that t is \([\frac{-\sqrt{1297}-23}{8},\frac{\sqrt{1297}-23}{8}]\)

Thanks! :)

Let's rewrite this equation into standard form ax^2+bx+c > 0. We have

\(x^2+6x>16-5x+33\\ x^2+11x-49>0\)

We could probably try out some form of the quadratic formula, but let's take another approach. 

Let's first complete the square for x. We get

\(\left(x+\frac{11}{2}\right)^2-\frac{317}{4}>0\)

Now, we can conformtably isolate x. We get

\(\left(x+\frac{11}{2}\right)^2>\frac{317}{4}\)

Since we square root it, we have to split the inequality into two different ones. We have

\(x+\frac{11}{2}<-\sqrt{\frac{317}{4}}\quad \mathrm{or}\quad \:x+\frac{11}{2}>\sqrt{\frac{317}{4}}\)

\(x<\frac{-\sqrt{317}-11}{2}\quad \mathrm{or}\quad \:x>\frac{\sqrt{317}-11}{2}\)

This is not the prettiest case, but it does work. Now, let's put this in interval notation. We have

\((-\infty, \frac{-\sqrt{317}-11}{2} )U(\frac{\sqrt{317}-11}{2}, \infty)\)

Thanks! :)

mmm...let's note something really quickly. 

In the problem, it said that

\(QR + PR < PQ \)

However, it said that PQR is a triangle, but this violates the Triangle Inequality Theorem.

So this problem is invalid. 

thanks! :)