NotThatSmart

Let's first take a look at the fraction 1/17. 

It has 16 repeating digits in the form of \(0. \overline{0588235294117647}\)

We simplfy have to find which digit the 4000th lies on. 

First, let's do 4000/16. 

We get

\(4000/16=250\)

So the 4000th digit is the last digit of the repeating decimal, which is 7. 

So 7 is our answer. 

Thanks! :)

Let's define some variables to help solve this problem.

Let's let \(BD=3-x\) and \(CD=x\)

From the problem and requirements given, we can write the equation

\(BD / AB = CD / AC \)

Plugging in our values from the problem and x, we have the equation

\((3 - x) / 4 = x / 5 \\ 5 (3 - x) = 4x \\ 15 - 5x = 4x \\ 15 = 9x\\ x = 15/9 = 5/3 = CD\)

Now, we can find the area of ADC easily! We have

\([ ADC ] = (1/2) (CD) ( AB) = (1/2) ( 5/3) ( 4) = 10 / 3 \)

So 10/3 is our answer. 

Thanks! :)

Height of ABC  =  AM

Base of ABC = BC

\([ ABC ]  = (1/2) (BC) (AM) =  14   → (BC) (AM)  =  28 \)

Height of BEC   = (1/2)AM

Base of BEC =  BC

\([ BEC ]  = (1/2) (BC) (1/2)AM =  (1/4)(BC) (AM)  =  (1/4) (28)  =  7\)

So our final answer is 7. 

Thanks! :)

Let's make some observations about the problem. Note that

\(CD = BD = 10 \\ ED = ED\)

From this, we can identify that triangle CDE and triangle BDE are congruent triangles.

This infromation tells us that \(BE=CE\)

Now, using the this new information, we can write the equation

\(sin CED / CD = sin CDE / CE \\ sin 75 / 10 = sin 90 / CE \\ CE = 10 sin 75 = BE ≈ 10.35\)

So our answer is about 10.35

Thanks! :)

We can note something really important to solve this problem. 

First off, let's note that

\((x+y)^3 = x^{3}+3x^{2}y+3xy^{2}+y^{3}\)

Plugging in all the information we already have from the problem, we get that

\(8=5+6xy\\ xy=1/2\)

The reason we need the value of xy will come into play later. 

Now, let's also note that

\((x+y)^2 = x^2+2xy+y^2\)

Isolating x^2+y^2, we get that

\(x^2+y^2 = (x+y)^2 - 2xy\)

We already know all the terms of the equation we needed to find x^2 + y^2. 

Plugging in 1/2 and 2, we get

\(x^2 + y^2 = 4 - 1 = 3\)

So our answer is 3. 

Thanks! :)

We can solve this problem by noting something extremely important. Note that we have

\(CD = BD = 10 \\ ED = ED\)

This is important because we now know that triangle CDE and triangle BDE are indeed congruent. 

From the congruence, we can find that \(BE=CE\)

Now, it's a breeze from here. We can write the equation

\(sin CED / CD = sin CDE / CE \\ sin 75 / 10 = sin 90 / CE \\ CE = 10 sin 75 = BE ≈ 10.35\)

So 10.35 is about our answer. 

Thanks! :)

We have the equation \((a+b)\sqrt{cd}\)

We have all the dimensions for us to solve the problem. 

Plugging in all the numbers given, we get

\((2+4)\sqrt{3(12)}\\ =6\sqrt{36}\\ =6(6)\\ =36\)

So 36 is our final answer.

Thanks! :)

From the problem itself, we can identify that

\(CD = BD = 10 \\ ED = ED\)

These equal sides prove to us that triangles CDE and BDE are congruent triangles. 

This means that \(BE = CE\)

Thus, we can write the equation

\(sin CED / CD = sin CDE / CE \\ sin 75 / 10 = sin 90 / CE \\ CE = 10 sin 75 = BE ≈ 10.35\)

So our answer is about 10.35 

Thanks! :)

I see what you mean now. 

Instead of it being \(4\choose 2\), we have overcounting to consider. 

Now, we only have 3 cases since there is no distinguishbility for the boxes. 

Let's name the balls ball 1, 2, 3, and 4. 

We have 

1/2 and 3/4

1/3 and 2/4

1/4 and 2/3

We already accounted for anyother case. 

Thank you for the correction, MEMEGOD!

Thanks! :)