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The answer is \(3*7=21\)

so 1 is the answer

The answer is \(1*2*6 \equiv 5 \pmod7\)

so 5 is the answer

In order to solve this problem, we can use a very simple trick. 

First, let's set \(x = 0.\overline{2123}\)

If we have this for the value of x, we get \(10000 x = 2123.\overline{2123}\)

Now, we subtract x from 10000x. Note that the repeating parts get canceled out. We get

\(10000x-x= 2123.\overline{2123} -\overline{2123}\\ 9999x=2123\)

This means we get

\(x = \frac{2123}{9999}\)

Simplifying, we get

\(x = \frac{193}{909}\)

So our final answer is \( \frac{193}{909}\)

Thanks! :)

Let's make some observations. 

Starting with \(2!\), every other term after that is even, since it contains 2. 

Since we know that \(even + even = even\), then we have \(2! + 3! + \dots + 100! = even\)

Thus, the sequence \(2! + 3! + \dots + 100! \equiv 0 (\mod 2)\)

However, we forgot about 1 term. Since , and 1 is odd, and we know , then we have

\(1! + 2! + 3! + \dots + 100! \equiv 1(\mod 2)\)

Thus, the answer is 1. 

Thanks! :)

First, let's put 0.11012 as a fraction. We have

\(\frac{0.11012}{1}\times \frac{100000}{100000}= \frac{11012}{100000}\)

Now, we must find a common factor between the two. Well, 4 is one, so dividing both the top and bottom by 4, we get

\(\frac{11012 \div 4}{100000 \div 4}= \frac{2753}{25000}\)

Unforutnately,  there's no more factors after this, so our final answer is 

\(0.11012 = \frac{2753}{25000}\)

Thanks! :)

mmm...take a look at the points given. 

\(P_1 P_2 + P_2 P_3 + P_3 P_4 + \dots + P_9 P_{10} + P_{10} P_1\)

This is essentially just the perimeter of the decagon \(P_1P_2P_3P_4P_5...P_{10}\)

The lenngth of one side is essentially

\( radius/2 ( -1 +\sqrt 5)\)

Now that we know one side, we can find the perimeter through the equation

\( 10 (1/2) ( -1 + \sqrt 5) = 5 ( -1 + \sqrt 5) ≈ 6.18\)

Thus, the answer is 6.18, 

Thanks! :)

"Hey, it's still possible to do it without the function" - famous last words 😂

Thanks, CPhill!

~NTS

Excellent, NTS !!!!

{Without the Totient Function , this would be very tedious....  Euler was a pretty smart guy......!!! }

https://web2.0calc.com/questions/probability_27845

Euler's Totient Function! The savior!

Let's look at 1200 first.

First, let me explain the process. 

We must prime factorize every number first. For 1200, we have

\(1200 = 12 * 5^2 * 2^2 = 2^4 * 3 * 5^2\)

Now, to summarize what we do, we essentially

\(\frac{\text{number we start with}}{\text{product of the prime factors}} \cdot \text{product of every prime factor -1}\)

So for 1200, we have \(\phi{1200} = 1200 / ( 2 * 3 * 5) * (2-1)(3-1) (5-1) = 40 (1)(2)(4) = 320\)

Meaning the answer is 320. 

Now, let's do the same for the other two numbers. 

First off, 181. It's a prime number, so the only prime factor is 181. Thus, we have

\(\phi 181 = \frac{181}{181}\cdot (181-1) = 180\). So 180 is the answer. 

Now, let's also do 212. We have that \(212 = 2^2*53\)

Thus, using Euler's Totient Function, we have

\(\phi212 = \frac{212}{2*53} \cdot (2-1)(53-1) = 2*52 = 104\) so 104 is the answer. 

Thus, the 3 answers are

\(\phi1200 = 320\\ \phi 181 = 180\\ \phi 212 = 104\)

Thanks! :)

Simplifying, we have

 sqrt [ ( sqrt 5 - sqrt 2)^2 ] + sqrt [ (sqrt 5 + sqrt 2)^2 ]

___________________________________________   =

                sqrt [ 5 - 2 ]

[ sqrt 5 - sqrt 2  ] + [ sqrt 5 + sqrt 2 ]

____________________________     =

                 sqrt 3

2sqrt 5

______  =

  sqrt 3

(2/3) sqrt 15

f(x) =  x^3 -6x^2 + 2x 

f (f(x)) =  [ x^3 - 6x^2 + 2x]^3 - 6 [ x^3 -6x^2 + 2x]^2 + 2[ x^3 - 6x^2 + 2x] =

x^9 - 18 x^8 + 114 x^7 - 294 x^6 + 300 x^5 - 312 x^4 + 154 x^3 - 36 x^2 + 4 x

So

x^3 -6x^2 + 2x  = x^9 - 18 x^8 + 114 x^7 - 294 x^6 + 300 x^5 - 312 x^4 + 154 x^3 - 36 x^2 + 4 x

x^9 - 18 x^8 + 114 x^7 - 294 x^6 + 300 x^5 - 312 x^4 + 153 x^3 - 30 x^2 + 2 x =  0

With a little help from WolframAlpha :

Solutions are :

Let's just count. 

We have

\((1,1) \\\to (13, 19)\\(\text{2 steps in between})\\\to (1, 19)\\\text{(3 steps in between)}\\\to(1, 3)\\\to(15,23)\\ \text{(2 steps in between)}\\\to(3, 23)\\\text{(4 steps in between)}\\\to(3, 3)\\\to (17,25)\\\text{(3 steps in between)}\\\to (1, 25)\\\text{(5 steps in between)}\\\to(1, 1)\)

Adding all of these up, we find that there are 28 points. 

Thanks! :)

The answer is \(244224 \pmod {64} \equiv 0\)

mmm...good question. I can't answer this question completely yet, but can you tell me what types of robots. 

One robot has the board Cytron Maker Rasberry Pi, which is for younger people

Here's a link on A* path finding

https://en.wikipedia.org/wiki/A*_search_algorithm

Here' what ai said

Good luck reading it...sorry for the small font. 

Now, I'll elaborate later, but I'm m,ore of a math guy....

Thanks! :)

Agreeable. sometimes the questions legit take me like 1-3 hours. (aka Like more than 20 entries for the answer in which are wrong) other times it takes me like 5-6 minutes. so I really don't know if I've got a hang of C and P yet or not 😂😂😂

Ty for giving hints only, I swear to god, last time it seemed that only blackjacked could read. Oh mannnnnnn Some of my friends, athey don't know like the last question and just search for answers. like i get it's the last question and Yall like took an hour but still haven't gotten it and all would be frustrating but, GOD DARN DO NOT ASK FOR ANSWERS. LIKE YALL GONNA LEARN AIR IF U ASK FOR ANSWERS. (and im not joking, dear friends who may or may not read this. You will actually learn air.)

                    C

            E

A              F          D          B

       7            2

Let DB = x

Since EF parallel to CD....Triangle AEF is similar to triangle ACD

AE/AC = AF/AD

AE/AC = 7/9

Since ED parallel to BC......Triangle AED is similar to triangle ACB

AE/AC = AD/AB

AE/AC = 9/ (9 + x)

So

7/9 = 9 / (9+x)

7(9+x) = 9*9

63 + 7x  = 81

7x = 18

x = 18/7  = DB

Using the Law of Cosines twice :

[BC^2 - BA^2 - AC^2] / [ -2 BA * AC]  = cos DAE

[8^2 - 3^2 - 7^2 ] / [ -2 * 3 * 7 ] =  cos DAE = -1/7

And

DE^2  = DA^2 + AE^2 - [2* DA * AE] cos ( DAE)

DE^2  = 21^2 + 9^2  - [ 2* 21 * 9] (-1/7)

DE^2  =  522 + 54

DE^2 = 576

DE = sqrt 576  =  24

191 is prime

phi (191)  =  (191 - 1)  =  190

Alright. I'm not the best with these problems, but let me give it a shot, 

First, let's use the Law of Cosines. According to the law, we have

\(BC^2 = AB^2 + AC^2 - 2(AB* AC) cos ( BAC)\)

Now, we know a lot of the information given already, so we want to find cos BAC. Thus, we have

\(8^2=3^2+7^2-2(3*7) \cdot cos(BAC)\)

Now, isolating BAC, we have that

\(64=16 \cdot cos(BAC)\\ cos(BAC)=4\)

Now, why not apply Law of Cosines again? Let's do it! :)

This time, let's use the equation

\(DE^2 = AE^2 + AD^2 - 2(AE * AD) cos (BAC) \)

We figured out and know every term in the right side of the equation, so plugging in all the information, we have

\(DE^2 = 9^2+21^2-2(9*21)*4\)

mmm...let's check something out.

This simplifies to \(DE^2=-990 \), which defintely IS NOT possible. 

Now, disclaimer though....it's kinda late at night, and my brain is fried. my solution is probably faulty. 

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