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What is the domain of the function
j(x) = 1/(x - 8) + 1/(x^2 - 8) + 1/(x^3 - 8)?

Hello Guest!

\(domain =\{x\in \mathbb R\}\)  |  {- ∞ < x < - \(\sqrt{8}\) }  {\(-\ \sqrt{8}\) < x < 2}    {2 < x < \(\sqrt{8}\) }    {\(\sqrt{8}\) < x < 8}     {8 < x < ∞}

  !

What is the radius of the circle?

Hello Guest!

\(A=\frac{\pi r^2}{360°}\cdot\alpha\\ r=\sqrt{\frac{360°\cdot A}{\pi\cdot \alpha}}\\ r=\sqrt{\frac{360°}{\pi\cdot 100°}\cdot \frac{250}{\pi}}=\dfrac{1}{\pi}\cdot \sqrt{36\cdot 25}\\ \color{blue}r=\dfrac{30}{\pi}=9.549\)

   !

235 ( 6 ) , 236 ( 6 ) , 255 ( 3 ) , 256 ( 6 ) , 266 ( 3 ) , 355 ( 3 ) , 356 ( 6 ) , 366 ( 3 ) , 555 ( 1 ) , 556 ( 3 ) , 566 ( 3 ) , 666 ( 1 ) , >>Total combinations = 12

>>Totals in parentheses = 44 permutations.

See https://web2.0calc.com/questions/functional-equation_6

Answer is (-7,6)

Let x = # of book marks Justin and Mike each have at first

        x - 20 = 2 (x - 33)

        x - 20 = 2x - 66

            46 = x

Let Ben's money = 2x

and John's money = y

Therefore x:y = 9:4

i.e x/y = 9/4 Cross multiply 

   4x = 9y -- (1)

After Ben spends $63

Ben's money becomes x - 63

      x - 63 : y = 1:2

       (x - 63)/y = 1/2

      2(x - 63) = y

      2x - 126 = y 

Input value of y in eq (1)

   4x = 9(2x - 126)

   4x = 18x - 1134

   1134 = 18x - 4x

   1134 = 14x

   1134/14 = x  x = 89

   Since 4x = 9y

     4 (81) = 9y

      324 = 9y

      324/9 = y

       36 = y

  Ben's money = $81 and

  John's money = $36

After Ben spends $63 his money

becomes $81 - $63 = $18 

John becomes $(36 - 18) more = $18 more

Initially Ben had 9x dollars and  John has 4x dollars.   

Ben spends  $63 so now   Ben has  9x-63   dollars and  John still has  4x dolllars.

This is in the ration 1:2

so

\(\frac{9x-63}{1}=\frac{4x}{2}\\ 9x-63=2x\\ 7x=63\\ x=9\)

You can finish it :)

Not my homework 

S = 1 / [1 - 2/7], solve for S

S==7/5

What is the remainder when the sum

\(1^2+2^2+3^2+...+2016^2 + 2017^2 + 2018^2 +2019^2 +2020^2\) 

is divided by 17?

\(\begin{array}{|rcll|} \hline 1^2 \pmod {17} &=& 1 \\ 2^2 \pmod {17} &=& 4 \\ 3^2 \pmod {17} &=& 9 \\ 4^2 \pmod {17} &=& 16 \\ 5^2 \pmod {17} &=& 8 \\ 6^2 \pmod {17} &=& 2 \\ 7^2 \pmod {17} &=& 15 \\ 8^2 \pmod {17} &=& 13 \\ 9^2 \pmod {17} &=& 13 \\ 10^2 \pmod {17} &=& 15 \\ 11^2 \pmod {17} &=& 2 \\ 12^2 \pmod {17} &=& 8 \\ 13^2 \pmod {17} &=& 16 \\ 14^2 \pmod {17} &=& 9 \\ 15^2 \pmod {17} &=& 4 \\ 16^2 \pmod {17} &=& 1 \\ 17^2 \pmod {17} &=& 0 \\ \hline \end{array}\)

cycle mod 17: \((1, 4, 9, 16, 8, 2, 15, 13, 13, 15, 2, 8, 16, 9, 4, 1, 0)\)

circle mod 17

\(\begin{array}{|rcll|} \hline && 1+ 4+ 9+ 16+ 8+ 2+ 15+ 13+ 13+ 15+ 2+ 8+ 16+ 9+ 4+ 1+ 0\pmod {17} \\ &=& 136 \pmod {17} \\ &=& \mathbf{0 \pmod {17}} \\ \hline \end{array}\)

2020/17 = 118 cycles remainder 14

remainder 14, this is the sum in the cycle from 1 to 14  mod 17

\(\begin{array}{|rcll|} \hline && 1+ 4+ 9+ 16+ 8+ 2+ 15+ 13+ 13+ 15+ 2+ 8+ 16+ 9\pmod {17} \\ &=& 131 \pmod {17} \\ &=&\mathbf{ 12 \pmod {17}} \\ \hline \end{array}\)

\( \begin{array}{|rcll|} \hline && 1^2+2^2+3^2+...+2016^2 + 2017^2 + 2018^2 +2019^2 +2020^2\pmod {17} \\ &=& \underbrace{0*118 \pmod {17}}_{118 \text{ cycles}} + \underbrace{\mathbf{ 12 \pmod {17}}}_{\text{remainder 14}} \\ &=& \mathbf{ 12 \pmod {17}} \\ \hline \end{array}\)

The remainder when the sum \(1^2+2^2+3^2+...+2016^2 + 2017^2 + 2018^2 +2019^2 +2020^2 \) is divided by 17 is \(\mathbf{12}\)

z^2 = 2i + 2.

let Z = a+bi

Z^2= a^2-b^2+2abi

\(a^2-b^2=2 \qquad (1)\\2ab=2\qquad (2)\\ ab=1\\ b=1/a\\ a^2-\frac{1}{a^2}=2\\ a^4-2a^2-1=0\\ let\;\;m=a^2\\ m^2-2m-1=0\\ ...\\ m=1+\sqrt2\qquad or \qquad m=1-\sqrt2\\ \text{but m is positive so}\\ m=1+\sqrt2\\ a=\pm\sqrt{1+\sqrt2}\\ b=\pm\frac{1}{\sqrt{1+\sqrt2}} \)

\(a=\pm\sqrt{1+\sqrt2}\\ b=\pm\frac{1}{\sqrt{1+\sqrt2}}\\ b=\pm\frac{1}{\sqrt{1+\sqrt2}}\frac{\sqrt{1+\sqrt2}}{\sqrt{1+\sqrt2}}\\ b=\pm\frac{\sqrt{1+\sqrt2}}{1+\sqrt2}\\ b=\pm\frac{\sqrt{1+\sqrt2}}{1+\sqrt2}\times\frac{1-\sqrt2}{1-\sqrt2}\\ b=\pm\frac{\sqrt{1+\sqrt2}}{1-2}\times\frac{\sqrt{1-\sqrt2}\sqrt{1-\sqrt2}}{1}\\ b=\pm\frac{\sqrt{1-2}}{1-2}\times\frac{\sqrt{1-\sqrt2}}{1}\\ b=\pm\frac{i}{-1}\times \frac{i\sqrt{\sqrt2-1}}{1}\\ b=\pm \sqrt{\sqrt2-1}\\ \)

You seriously need to check that

PLUS I expect it should ahave been done by letting     \(Z=Ae^{i\theta}\)

LaTex

a^2-b^2=2 \qquad (1)\\2ab=2\qquad (2)\\
ab=1\\
b=1/a\\
a^2-\frac{1}{a^2}=2\\
a^4-2a^2-1=0\\
let\;\;m=a^2\\
m^2-2m-1=0\\
...\\
m=1+\sqrt2\qquad or \qquad m=1-\sqrt2\\
\text{but m is positive so}\\
m=1+\sqrt2\\
a=\pm\sqrt{1+\sqrt2}\\
b=\pm\frac{1}{\sqrt{1+\sqrt2}}

a=\pm\sqrt{1+\sqrt2}\\
b=\pm\frac{1}{\sqrt{1+\sqrt2}}\\
b=\pm\frac{1}{\sqrt{1+\sqrt2}}\frac{\sqrt{1+\sqrt2}}{\sqrt{1+\sqrt2}}\\
b=\pm\frac{\sqrt{1+\sqrt2}}{1+\sqrt2}\\
b=\pm\frac{\sqrt{1+\sqrt2}}{1+\sqrt2}\times\frac{1-\sqrt2}{1-\sqrt2}\\
b=\pm\frac{\sqrt{1+\sqrt2}}{1-2}\times\frac{\sqrt{1-\sqrt2}\sqrt{1-\sqrt2}}{1}\\
b=\pm\frac{\sqrt{1-2}}{1-2}\times\frac{\sqrt{1-\sqrt2}}{1}\\
b=\pm\frac{i}{-1}\times \frac{i\sqrt{\sqrt2-1}}{1}\\
b=\pm \sqrt{\sqrt2-1}\\
 

Let x = Clay's money

Let y = Timmy's money

            x + y = 318 (1)

            y = x - 18 (2)

Substitute (2) to (1):

        x + x - 18 = 318

           2x/2 = 336/2

           x = 168

From (2): y = x - 18

                y = 168 - 18

                y = 156

Let z = amount of money Timmy spent at the store

(150 - z) = 168

900 - 6z = 168

732/6 = 6z/6

z = $122

Tommy had 192 stickers

Terrance had 116 stickers

Assume that they have given 'x' number of stickers

Now, no. of stickers tommy had = 192 - x

         no. of stickers terrance had = 116 - x

        Given that Tommy, now have 5 times as many stickers as Terrance

            192 x - = 5 (116 - x)

            192 - x = 580 - 5x

            5x - x = 580 - 192

              4x = 388

                x = 97

Tommy and Terrance had 97 stickers.

How can 50 votes be distributed across 4 people

So divide the 50 votes by 3 bars.

The first canditate gets what is in front of the first bar

the second gets what is infront of the 2nd bar

the third gets waht ias in front of the 3rd bar

the fourth gets what is after the 3rd bar

53C3=23426

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Next time please present your question problerly, those meaningless dollar signs are annoying.

I am less likely to answer annoying questions.

AND I just realized that you down voted guests incorrect answer. 

THAT IS NOT NICE NOR WAS IT CALLED FOR.

They were trying too help you by giving you a suggestion that they thought was correct and you punished them for it.  You should have thanked them for showing an interest in your question.

You can finish it

I hope this is correct.

The difference between A and B is (12, 6). Since BC is 1/4th of AB, the difference between B anc C is (3, 3/2). We add this to B to get (17, 11/2).