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p + q = 3 + (-17) = -14.

1. I think the answer is 10,000,000.

4. It's (2/3)^3 = 8/27.

Solution: 

This method uses Euler's totient and the modulo inverse functions.

\(\begin{array}{rcll} n &\equiv& {\color{red}1} \pmod {{\color{green}4}} \\ n &\equiv& {\color{red}1} \pmod {{\color{green}3}} \\ n &\equiv& {\color{red}2} \pmod {{\color{green}5}} \\ \text{Set } m &=& 4\cdot 3\cdot 5 = 60 \\ \end{array} \)

\(\small{ \begin{array}{l} n = {\color{red}1} \cdot {\color{green}3\cdot 5} \cdot \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ [ { (\color{green}3 \cdot 5) }^{\varphi({\color{green}4}) -1 } \pmod {{\color{green}4}} ] }_{=\text{modulo inverse }(3\cdot 5) \mod 4 } }_{=(3\cdot 5)^{2-1} \mod {4}} }_{=(3\cdot 5)^{1} \mod {4}} }_{=(15\pmod{4})^{1} \mod {4}} }_{=(15)^{1} \mod {4}} }_{= 3} + {\color{red}1} \cdot {\color{green}4\cdot 5} \cdot \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ [ { (\color{green}4\cdot 5) }^{\varphi({\color{green}3}) -1} \pmod {{\color{green}3}} ] }_{=\text{modulo inverse } (4\cdot 5) \mod {3}} }_{=(4\cdot 5)^{2-1} \mod {3}} }_{=(4\cdot 5)^{1} \mod {3}} }_{=(20\pmod{3})^{1} \mod {3}} }_{=(2)^{1} \mod {3}} }_{=2} + {\color{red}2} \cdot {\color{green}4\cdot 3} \cdot \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ [ { (\color{green}4\cdot 3) }^{\varphi({\color{green} 5}) -1 } \pmod {{\color{green}5}} ] }_{=\text{modulo inverse } (4\cdot 3) \mod 5 } }_{=(4\cdot 3)^{4-1} \mod { 5}} }_{=(4\cdot 3)^{3} \mod {5}} }_{=(12\pmod{5})^{3} \mod {5}} }_{=(2)^{3} \mod {5}} }_{=3}\\ \\ n = {\color{red}1} \cdot {\color{green}3\cdot 5} \cdot [3] + {\color{red}1} \cdot {\color{green}4\cdot 5} \cdot [2] + {\color{red}2} \cdot {\color{green}4\cdot 3} \cdot [3] \\ n = 45+ 40 + 72 \\ n = 157 \\\\ n \pmod {m} = 157 \pmod {60} \\ n = 37+ k\cdot 60 \; \; k \in Z\\ \mathbf{n_{min}} \; \mathbf{=} \; \mathbf{37} \end{array} }\)

GA

I'm getting z = 2 - 7i.

I finally got it. The answer is 4/1. Turns out I was multiplying the wrong numbers in the wrong mode :D Thank you so much for being so thorough!!!

(a) We can choose four numbers from 1, 2, 3, ..., 12.  Then a is the lowest, b is the second-lowest, c is the third-lowest, and is the fourth-lowest, so the number of ways of choosing a,b,c,d is C(12,4) = 495.

(b) There are 12 ways of choosing a.  If a = 12, then there are 12 ways of choosing b.  If a = 11, then there are 11 ways of choosing b.  This pattern continues, so the number of ways of choosing b is 12 + 11 + 10 + ... + 1.

If b = 11, then there are 11 ways of choosing c.  If b = 10, then there are 10 ways of choosing c.  This pattern continues, so the number of ways of choosing c is 11 + 10 + 9 + ... + 1.

If c = 10, then there are 10 ways of choosing d.  If c = 9, then there are 9 ways of choosing d.  This pattern continues, so the number of ways of choosing d is 10 + 9 + 8 + ... + 1.

So the number of ways of choosing a,b,c,d is 12(12 + 11 + ... + 1)(11 + 10 + 9 + .. + 1)(10 + 9 + 8 + ... + 1) = 12*78*66*55 = 3397680.

(c) There are n ways of choosing a_1.  If a_1 = n, then there are n ways of choosing a_2.  If a_1 = n - 1, then there are n - 1 ways of choosing a_2.  This pattern continues, so the number of ways of choosing a_2 is n + (n - 1) + ... + 1 = n(n + 1)/2.

We can take the formula for part (b) and make it for n and k.  The number of ways of choosing a_1,a_2,...,a_k is n * n(n + 1)/2 * (n - 1)n/2 * ... * (n - k)(n - k + 1)/2.

(a) We can choose four numbers from 1, 2, 3, ..., 12.  Then a is the lowest, b is the second-lowest, c is the third-lowest, and is the fourth-lowest, so the number of ways of choosing a,b,c,d is C(12,4) = 495.

(b) There are 12 ways of choosing a.  If a = 12, then there are 12 ways of choosing b.  If a = 11, then there are 11 ways of choosing b.  This pattern continues, so the number of ways of choosing b is 12 + 11 + 10 + ... + 1.

If b = 11, then there are 11 ways of choosing c.  If b = 10, then there are 10 ways of choosing c.  This pattern continues, so the number of ways of choosing c is 11 + 10 + 9 + ... + 1.

If c = 10, then there are 10 ways of choosing d.  If c = 9, then there are 9 ways of choosing d.  This pattern continues, so the number of ways of choosing d is 10 + 9 + 8 + ... + 1.

So the number of ways of choosing a,b,c,d is 12(12 + 11 + ... + 1)(11 + 10 + 9 + .. + 1)(10 + 9 + 8 + ... + 1) = 12*78*66*55 = 3397680.

(c) There are n ways of choosing a_1.  If a_1 = n, then there are n ways of choosing a_2.  If a_1 = n - 1, then there are n - 1 ways of choosing a_2.  This pattern continues, so the number of ways of choosing a_2 is n + (n - 1) + ... + 1 = n(n + 1)/2.

We can take the formula for part (b) and make it for n and k.  The number of ways of choosing a_1,a_2,...,a_k is n * n(n + 1)/2 * (n - 1)n/2 * ... * (n - k)(n - k + 1)/2.

(a) We can choose four numbers from 1, 2, 3, ..., 12.  Then a is the lowest, b is the second-lowest, c is the third-lowest, and is the fourth-lowest, so the number of ways of choosing a,b,c,d is C(12,4) = 495.

(b) There are 12 ways of choosing a.  If a = 12, then there are 12 ways of choosing b.  If a = 11, then there are 11 ways of choosing b.  This pattern continues, so the number of ways of choosing b is 12 + 11 + 10 + ... + 1.

If b = 11, then there are 11 ways of choosing c.  If b = 10, then there are 10 ways of choosing c.  This pattern continues, so the number of ways of choosing c is 11 + 10 + 9 + ... + 1.

If c = 10, then there are 10 ways of choosing d.  If c = 9, then there are 9 ways of choosing d.  This pattern continues, so the number of ways of choosing d is 10 + 9 + 8 + ... + 1.

So the number of ways of choosing a,b,c,d is 12(12 + 11 + ... + 1)(11 + 10 + 9 + .. + 1)(10 + 9 + 8 + ... + 1) = 12*78*66*55 = 3397680.

(c) There are n ways of choosing a_1.  If a_1 = n, then there are n ways of choosing a_2.  If a_1 = n - 1, then there are n - 1 ways of choosing a_2.  This pattern continues, so the number of ways of choosing a_2 is n + (n - 1) + ... + 1 = n(n + 1)/2.

We can take the formula for part (b) and make it for n and k.  The number of ways of choosing a_1,a_2,...,a_k is n * n(n + 1)/2 * (n - 1)n/2 * ... * (n - k)(n - k + 1)/2.

Find the length of an arc on a circle of radius 8 corresponding to an angle of (90/π)∘

I keep getting the answer of 229.12 but it asks me to enter it as a fraction

When you get answers Roxettna you should learn to look at whether that answer is reasonable.

Here you have a circle of radius 8 so the circumference is 2*pi*8 which is about 50units.

pi is just a bit more than 3 so 90/pi is about 30degrees.  Which is only a little bit of a circle so the answer must be way less than 50.

Your answer of 229 could not possibly be correct!

Now  the circumference of a circle is    2*pi* r

There are 360 degrees in a revolution.

so

you have     90/pi  divided by 360   of the circle

So

\(Arc\; length = \frac{\frac{90}{\pi}}{360}*2\pi r\\ Arc\; length = (\frac{90}{\pi}\div 360)*2\pi r\\ Arc\; length = (\frac{90}{\pi}\times \frac{1}{360})*2\pi r\\ Arc\; length = \frac{1}{4\pi}*2\pi r\\\)

You should be able to finish it from there. Don't forget that r is 8 units    

M, as in macabresubwoofer, does not get help. At least not to learn.

It is more than evident that Web2 is just a cheat site for macabresubwoofer.

And yes, he (possibly she) is abusive to the people who do his homework for him.

Idk, I found this : https://brainly.com/question/5774117

Looks complicated solution.. (No value) 

wait why didn't anybody respond,,,,

Looking at his profile history, yeah his questions seem a little too easy.

"Yes, first of all i do my own homework... it's just only a few questions i can't seem to get quite easily" - Macabres

That sounds quite skeptical based on the questions you ask.

But hey? Who is to judge the age or intelligence of Macabres?

awesome!! :D 

Yup just edited and replied!

Believe it or not,,, i was just thinking of that too...

do you know what language you're gonna learn CU? :D 

Thanks, probs during the summer (which is far far away lol) I am going to take some computing classes so I can be cool like you 

Probably I will start with Java, because it  is the most common and easy one.

THank you Guest and Nirvana!

you beat me to it :') 

i was about to say it's not python i think its C++ 

but dang!! C++ is a really hard language to learn, congrats :) 

It is written in C++.

they didn't seem to use the semicolon at the end of their lines of code...so i'm guessing python (don't take my word, but try to run it through to see if it works)

it could also be something funky as C++ or C# or something which i have NOOOO idea how that works lol 

Also, what program do you use to run the code? I would like to use your codes to help me solve these questions in the future quickly!

Sorry, I missed that too !!

So, 72 + 66 =138 members  in the band.

OMG THANK YOU SO MUCH! I just realized that the numbers are required to be relatively prime!
 

Thats why I got the problem wrong when I followed the video, because the Question had no relatively prime numbers! The question was making you guess and check.

I can solve system of modulos now! Thanks!

Oops! Between 100 and 200.

A lesson learned... read the question carefully! 

N mod 8 = 2

N mod 9 = 3

N =72 m +  66, where m=9, 1, 2, 3.........etc.

There are 66 members in the band.