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This computer code incorporates the "Chinese Remainde Theorem + Modular multiplicative Inverse"
i=0;j=0;m=0;t=0;a=(4, 3, 5);r= (1, 1, 2);c=lcm(a); d=c / a[i];n=d % a[i] ;loop1:m++; if(n*m % a[i] ==1, goto loop, goto loop1);loop:s=(c/a[i]*r[j]*m);i++;j++;t=t+s;m=0;if(i< count a, goto4,m=m);printc,"m + ",t % c;return

ANSWER =60 m +  37, where m=0, 1, 2, 3......etc.

Look under this link for detailed explanation of CRT:
https://web2.0calc.com/questions/congruences

Look at guests answer below LOL I need more modulo practice

well, I guess you are welcome! 

let me just tell you CU, you got all of it right...now that i tried out what you said 

:)

except for the hexagon part: i'm a little confused....

Not sure our class has just touched the surface of constructions. we just started constructing congruent triangles lol very basic

1. I think you can prove the segment is bisected by showing that it is divided into TWO EQUAL lengths, so probably you have to use your compass and make two swoops that are equal in swoop-length, I don't know.

If a segment is divided into two congruent parts, then it is bisected.

2. I think you just have to show that they form 90 degrees, and that like number 1, it divides the segment into TWO EQUAL parts.

3.

Since all radii of a circle are congruent, you probably have to show that the triangles of the hexagon meet at the center.

If you do this, you proved two of their corrresponding sides are congruent. Then using vertical angles, you can prove all of the included angles of their sides are congruent.

So by doing this, you can use SAS to show that ALL of the triangles are congruent.

Then you have to show that ONE of the triangles is equilateral, which makes ALL of the other triangles equilateral as you just proved them all to be congruent

A triangle is equiliateral if and only if all of its corresponding sides are equal OR if all of its corresponding angles are equal.

Sorry, what I said is quite confusing, proofs are hard.

I haven't learned construction proving quite yet, so don't take my answer seriously.

and for sure you will CU! 

i know you will :D

 _{the first is 1/3 of the sum of the term that precedes it and the term that follows it in the sequence}

2, x, y, 34

That is the sequence we have so far

Lets model an equation after that:

\(\frac{1}{3}(2+y)=x\)

\(\frac{1}{3}(x+34)=y\)

solve(1/3(2+y)=x, (1\3)(x+34)=y = {y=13, x=5}

Solving, we get the 3rd term being equal to 13.

Knowing that the third term is 13, the fourth term is 34, we can calculate the 5th term as follows:
 

\(\frac{1}{3}(13+x)=34\)

With x being the 5th term

(13+x)/3=34 = x=89

\(\boxed{89}\)

I haven't taken Alg 2 yet, but I can attempt it.

Imaginary number is \(\sqrt{-1}\)

So

\(w=3+\sqrt{-4}\)

Plugging in \(|w+z|=13\)

We have

\(3+\sqrt{-4}+z=13\)

\(\sqrt{-4}+z=10\)

We have

\(z=10-\sqrt{-4}\)

\(z=10-4i\)

Does this even work??

........ I mean what I said 4i

4i is a complex number....

if you you do not know what that is then you may as well not bother attempting this problem as you need an understanding of magnitude and complex numbers to solve it.

Organize as follows

\((x^2+y^2)+(xy+x)+(y+1)\)

Factor as follows 

\((x^2+y^2)+x(y+1)+1(y+1)\)

Re-organize as follows

\((x^2+y^2)+(x+1)(y+1)\)

We see that negative values for x and y won't make it any more minimum

We instantly know that substituting 0 will get the smallest value.

(0+0) + (1)(1)

1

The minimum value is \(\boxed{1}\)

Thanks guest for confirming answer, I just did it the complicated way so we can be sure that negative values don't decrease the minimum value

The minimum value is 1.  Plug in x = y = 0.

what do you mean by 4i? 41 or you actually mean 4i?

This guest was always saying these things when M got help... 

Wonder if he ever got tired of posting. 

Use aops or try to find tricky problems no one has really seen and try to work them out

Can  you insert the picture?

Sorry the picture is blocked on my computer

I requested access dunno if it will work.

After rotating my head 450 degrees and looking at EFGH, I realized I could form a rectangle by moving triangle EFN with its side EF to match up to HG.

Don't worry my neck is fine

Wait I am so sorry. The pictures won't upload.

yea ok 

The order doesn't really matter, except for Dr. Jones and Dr. Smith.

Hmmmm, this seems to be a case-work problem

Since Dr. Smith MUST be after Dr. Jones,

he can be the 2nd, 3rd, 4th, 5th, or 6th.

So try to find the amount of lectures for 

2nd

3rd

4th

5th

6th

then add em up. I think thats what you do.

Right... exactly what you mean and by the way i don't mean to show anyone that use anyone or seem lazy to others on this site like i don't do my work but i do and thanks for defending me on that too

i dunno he is your average toxic guest

I personally have no problem with it, but you might have to justify your question legitimacy in order to not recieve fire.

guests are toxic because they are anonymous, allowing them to say whatever they want without getting accused.

Lol it still doesn't work.

Can you copy paste?

If you can't try sending screen shots