NotThatSmart

We can use simple long division to solve this problem.

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! :)

We can apply a handy trick to solve this problem. 

Let's set a variable. Let's set \( x=0.29\overline{6}\)

We want to put x in terms of a fraction. 

Now, for that value of x, we have \(10 x = 2.96\overline{6}\)

This is important, as now we subtract x from 10x. We get

\(10x-x = 2.96\overline{6}-0.29\overline{6}\)

Since the repeating decimal cancels out, we have

\(9 x = 2.67\). 

Now, we simply solve for x. We have

\(x = \frac{2.67}{9}\)

\(x=\frac{2.67}{9}\times \frac{100}{100}= \frac{267}{900}\)

\(x = \frac{89}{300}\)

Thus, our final answer is 89/300. 

Thanks! :)

We can solve this problem by defining variables and using a nice little trick. 

First, let's set \(x = 0.031\overline{4}\). we are trying to put x in fractional form. 

For that value of x, we have that \(10 x = 0.314\overline{4}\)

Now, we subtract x from 10x. We get

\(10x-x = 0.314\overline{4}-0.031\overline{4}\)

Since the repeating 4 cancels out, we are left with

\(9 x = 0.283\)

Now, we simplfy solve for x. 

\(x = \frac{0.283}{9}\)

\(x=\frac{0.283}{9}\times \frac{1000}{1000}= \frac{283}{9000}\)

So, we have

\(0.031\overline{4} = \frac{283}{9000}\)

So 283/9000 is our final answer. 

Thanks! :)

In order to convert this to bae 10, we use a simple trick. 

We multiply every digit in 11021 by a power of 9 increasing from 0. 

So, our answer would be

\((11012)_9 = (1 × 9^4) + (1 × 9^3) + (0 × 9^2) + (1 × 9^1) + (2 × 9^0) = (7301)_{10}\)

We multiply the units digit by 9^0 and increase from there on. 

So, our final answer is 7301. 

Thanks! :)

First, let's expand \((x+y)^2\) and see what we get. 

We have \((x+y)^2 = x^2+2xy+y^2\) as our full expansion. 

Setting \((x + y)^2=x^2 + y^2\), we have the equation

\(x^2+2xy+y^2 = x^2+y^2\). Cancelling out terms, we get that

\(2xy=0\\ xy=0\)

So we must have xy=0, in order for it to be possible. 

This means either x or y must be 0 for this to happen. If neither are 0, then we have the equation at false. 

So our answer is \(xy=0\)

Thanks! :)

In order to solve this problem, we need to find how many ways we can arrange each age group. 

The freshmen can stand in a line \(4! = 24\) ways. 

As a group, they can arrange themselves in 4 ways across the line. 

The sophoores can stand in a line \(3! = 6 \) ways. 

Thus, we simply do

\(24 * 4 * 6 = 576 \)

So 576 is our final answer. 

Thanks! :)

We already have all the information we need to find \(\sqrt[3]{2a^2 b^4} + \dfrac{a - c}{(b+c)^2}\), so let's plug them in. 

We get

\(\sqrt[3]{2(4)^2 (2)^4} + \dfrac{4 +5}{(2-5)^2}\)

Simplifying and multiplying everything out, we get

\(\sqrt[3]{512} + \dfrac{9}{(-3)^2}\\ 8+1\\ 9\)

So our final answer is 9. 

I may have made a mistake with computation. If I did, let me know. 

Thanks! :)

First, let's convert 1/288 into base 25 form. 

We get that \((BD)_{25} = (11 × 25^1) + (13 × 25^0) = (288)_{10} \)

This means that \(\frac{1}{288}_{10} = \frac{1}{BD}_{25}\)

Doing base division, we have that

\(\frac{1}{BD}_{25} = 0.\overline{0.02468ACEGIKN}\)

So when \(\frac{1}{288}\) is in base 25, the decimal is repeating. 

So repeating is our answer. 

Thanks! :)

I made a slight mistake for this problem. 

We have \(1/17=0.\overline{0588235294117647}\). There are 16 repeating digits in this decimal. 

Thus, we must do \(4000/16=250\)

Since the number is divisble, the last digit is the 4000th digit. 

So 7 is our answer. 

Thanks! :)