All Answers 356328 Answers

We can use long division for this problem. 

The first 10 digits are \(13/29 = 0.4482758620\)

So our final answer is 0.4482758620. 

Thanks! :)

This problem was solved before. It should help you a lot. 

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! :)

Number of diaginals in a regular polygon =   (n) (n-3) / 2   ....  where n = the number of sides

So

(n) ( n -3)  / 2  =  2n

n^2 - 3n  = 4n

n^2 - 7n = 0

n ( n  - 7) = 0

n - 7 = 0

n = 7 

A = 6 (1/2) (side length)^2 * sqrt (3) / 2

3 = 3  (side length)^2 * sqrt (3) /2

2 /sqrt 3  = side length^2

sqrt [ 2 / sqrt 3 ]   = side lengrh ≈  1.075

https://web2.0calc.com/questions/problem-on-spheres

https://web2.0calc.com/questions/help-geometry_57877

https://web2.0calc.com/questions/geometry_47534

6(12)^4 + 0 (12)^3 + 3(12)^2 + 2(12) + 9   =  124881₁₀

9(16)^2  + 8(16) + 7  = 2439₁₀

35 / 8 =  4 R 3

4 / 8  =   0 R 4

35₁₀  = 43₈

\(6\left( 1\cdot1 + 2\cdot2 + \dots + n(n) \right) \)

For the moment, disregard the "6"

We have

1 +  4 +  9 +  16  +  25 +  36

 1      5     14      30     55    91

    4      9     16      25     35

        5     7       9        11

            2      2        2

We have  three rows of non-zero differences

The generating polynomial willl be

an^3 + bn^2 + cn + d

We have this system

a + b + c + d  = 1

8a + 4b + 2c + d = 5

27a + 9b + 3c + d = 14

64a + 16b + 4 + d = 30

Solving this  produces

a = 1/3     b =  1/2  c  =1/6   d  = 0

The polynomial is

(1/3)n^3 + (1/2)n^2 + (1/6)n

Multiplying by 6  we get

f(n) = 2n^2 + 3n  + n

https://web2.0calc.com/questions/floor-function_46

xy + x = (3x + 2y + z) / 2

2 (xy + x)  = 3x + 2y + z

2xy + 2x =  3x + 2y + z

2xy - x = 2y + z

x ( 2y -1)  = 2y + z

x = ( 2y + z)  / ( 2y  -1)