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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! :)

Working backwards,

\(a_9 = 1 - 4 = -3 \)

\(a_8 = 4 - (-3) = 7\)

\(a_7 = -3-7=-10\)

\(a_6 = 7-(-10)=\boxed{17}\)

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! :)

take the first and last number. sum them 

\(140+1=141\)

multiply by half the number of total numbers. \(140/2=70\)

we thus have \(141*70=9870 \)

prime factorize 9870. \(9870 = 2 * 3 * 5 * 7 * 47\)

take largest prime number --> 47

47 is the answer. 

Hi EP! :)

\(1\% * 200 = 2\)            this means that only 2 kg were not water. 

let x be the new amount

\(2 = 10\%x\)

solve for x

\(2(10)=x\\ x=20\)

20kg is the answer. 

The answer is \(0\)

The answer is \(1\)

\(1^2\equiv1\pmod{11}\)

The answers are \(2,4\)

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! :)

As posted yesterday:

(n+2) / n    *   ( n+3)/ (n+1)  *  (n+4) / (n+2) = 10      <==== Given 

Simplifies to 

1/n * (n+3) / (n+1) * (n+4) = 10 

then:

(n^2 + 7n + 12 )/ (n^2 +n) = 10        doing the fraction division results in:

9n^2 + 3n-12 = 0 

Use Quadratic Formula with    a = 9    b= 3    and c = -12     to find  n = 1      (or - 4/3  <=== does not work) 

So the first fraction would be 3/1 

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! :)

First off, since both of them equal to y, then we can set the two equations to equal each other. We have

\(\log2 (2x) =\log4 (16 + x)\)

Applying the change of base thereom to the equation, we get

\(\log (2x) / \log2 = \log (16 + x) / \log 4 \\ \log (2x) / \log 2 = \log (16 + x) /\log 2^2 \\ \log (2x) / \log 2 = \log (16 + x) / [ 2 \log 2]\)

Now, we multiply through by log 2. 

\(\log (2x) =\log (16 + x) / 2 \\ 2\log (2x) = \log (16 + x) \\ \log (2x)^2 = \log (16 + x)\)

Now, wait a second. Notice that both logs have the same base now. 

This means there aren't any reprucussions we have to worry about. We simply just take the inside of the parenthesis. 

This implies that

\((2x)^2 = 16 + x \\ 4x^2 - x - 16 = 0\)

Taking the postiive value of x and using the quadratic equation, we get

\(x = [ 1 + \sqrt{257} ] / 8\)

The answer to the system of equations is \(x = [ 1 + \sqrt{257} ] / 8\)

First, find the slope as  (y1-y2) /( x1-x2) = (8-17)/(5 - -4) =  (-9)/(9) = - 1

Then the line in mx+b form is 

    y = (-1) x + b     sub in the values of either point to find b = 13

so the equation becomes 

    y = -x + 13     re-arrange 

  x +y = 13      divide thru by 13 

  1/13x  +  1/13 y  =1 

I do not think this is a geometric series. 

We have 

\(0.0000324/0.036=0.0009\)

\(0.036/0.4=0.09\)

mmmm...🤔🤔🤔

I think this is correct....

Thanks! :)

Let's do a very sneaky trick to solve this problem. Now, if we add the two sequences together, we get

\(a_1 + a_3 + a_5 + a_7 + a_9 + a_{11} = 0 \\ a_2 + a_4 + a_6 + a_8 + a_{10} + a_{12} = 0 \space + \)

___________________________________

\(a_1+ a_2+ a_3+ \dots+ a_{10}+ a_{11}+ a_{12}=0\)

Now, this sum is really important. 

Now, first, let's consider the possibility of something where half the numbers are negative, and the other half is the absolute value of the numbers. 

Although this is possible, we have an even number of terms, meaning it's impossible to account for 0. 

EX: \(-5, -4, -3,-2,-1,0,1,2,3,4,5 (6 \text{ extra})\)

Thus, the only possibility is EVERY SINGLE TERM is 0. 

This satisfies everything, so meaning that \(a_1=0\)

So 0 is our answer. 

I hope I'm right...

Thanks! :)

Oh yeah..nice catch EP. I don't know what my brain was thinking. 

It was 9 in the morning, not sure if that's an excuse...LOL

Nice work EP. Sorry for the mistake. 

I like how I just switched between the two numbers randomly...

~NTS