NotThatSmart

There's a very simple solution to this. We have

\(AB = BC\)

Also note that Triangle AXB is similar to Triangle ABC

Because of that correlation, we can write

\(AX / AB  = AB / AC\\ 5 / AB = AB / 10\\ AB^2  = 50 \\ AB = 5\sqrt 2 = BC\)

Using the other sides of the triangles, we also know that

\(BX / AB  = BC / AC\\ BX / AB = AB  / AC \\ BX / AB = AB / 10\\ 10BX = AB^2\\ 10 BX = 50\\ BX = 50 / 10    =  5\\\)

Thus, our final answer is 5. 

Thanks! :)

Let's make some observations about the problem. 

First off, let's note that Triangle ABC is  a 45-45-90 right triangle

This means that we have \( AB = BC  =  12/\sqrt 2 =  6\sqrt 2\)

From the problem, we also have that

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

So triangles  ABD  and CBD  are congruent

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

Thanks! :)

We can use handy tricks to solve this problem. 

Let's focus on the last digit of the sum. This will come in handy. 

- 5 raised to any power will end in 5

7's last digit follows a pattern

\(7^1  = 7 \\ 7^2 = 49 \\ 7^3  = 343\\ 7^4 = 2401\)

- Since it repeats in groups of 4 7^13  ends in 7

\(5^19 + 7^13 + 23\)    will end in  \( 5 + 7 + 3 =  5 \)

This means 2 cannot be a divisor. 

The smallest  prime divisor wil either be 3  or 5

\([5^19 + 7^13 + 23] \pmod 3  = 2 \\ [5^19 + 7613 + 23 ] \pmod 5 = 0\)

The smallest prime divisor is 5

Thanks! :)

Since we can't use a calculator (thanks a lot LOL :))

Look at the image!

We have that \(2/17 = 0.\overline{1176470588235293}\)

There are 16 digits repeating in this decimal. 

So our answer is 1176470588235293. 

I'm not sure how there's only 6 digits, but there are 16 digit repeating. 

Thanks! :)

The smallest possible prime divisor a number can have is of course, 

If we check the multiplication, we can see that 3 odd numbers can't possibly mulitply to be an even. 

So, \(5^{19} * 7^{13} * 3^{31}\) is not divisble by 2. 

The next smallest is 3. 

The third term of the multiplication is \(3^{31}\)

This means, no matter what, this number has to be divisible by 3. 

So 3 is our final 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! :)

Let's give this problem a shot. 

Now, we know that \((a+b)^2 = a^2+2ab+b^2\)

From the problem, we ALSO know that \((a+b)^2 = 1^2=1\)

Thus, we have the equation \(a^2+2ab+b^2=1\)

Since the problem gives us the value of \(a^2+b^2=2\), we can easily figure out what ab is. 

We have

\(2ab+2=1\\ 2ab=-1\\ ab=-1/2\)

The value of ab will come in handy later. 

Now, let's focus on what we must find. a^3+b^3. From a handy equation, we know that

\(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)

Wait! We already know all the values needed to solve the problem. Plugging in 1, 2, and -1/2, we get

\(a^3+b^2=(1)(2-(-1/2)) = 2+1/2 = 5/2\)

Thus, our final answer is 5/2. 

Thanks! :)

*1300 points

We have

\(a \equiv 5 \pmod{7}\\ a+1 \equiv 5+1 \pmod7\\ a+1 \equiv 6 \pmod 7\)

Thus, the answer is 6. 

Testing numbers like 

\(5,12,19,26,33\), they all satisfy the equation we are given. 

Thanks! :)

Having a double root means that the descriminant of the quadratic is 0. 

Let's simplify the equation. 

Combining all like terms and simplifying, we have

\(x^2 + 15x + k \)

Now, since the descriminant is 0, we have the equation

\(15^2 - 4 (1)(k) = 0 \\ 225 - 4k = 0\\ 225 = 4k \\ k = 225 / 4 \)

Thus, our final answer is 225/4. 

Thanks! :)

First, let's make some observations about this question. 

All points 5 units away from point (2, 7) would make a circle with center (2, 7) and radius 5. 

The formula for that circle would look something like \((x-2)^2+(y-7)^2=25\)

Now, combining that with the second equation, and we get a system. 

Subsituting out y with the x values in the line equation, we get 

\((x-2)^2+(5x-35)^2=25\)

Simplifying, and we get

\(13x^{2}-177x+602=0\)

Using the quadratic equation, we find two values of x. We get

\(x=7\\ x=\frac{86}{13}\)

Plugging these x values into the second equation where we isolated y, we get that \(y=7;y=\frac{66}{13}\)

So our final two answers are (7, 7) and (86/13, 66/13)

Thanks! :)