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diagram 2

Your question was automatically  flagged because of the language that you used. It is not helpful in life to go around calling everything that you find difficult 'stupid'. The only think that it brings into question is your own intelligence.

Anyway I have unflagged it and will answer I hope you learn more from my answer than just how to cheat with minimum effort.

A proper thank you or polite feedback will be appreciated.

question

I will do a pictorial answer. There may be an easier way but then again the problem was not supposed to be easy.

I will also do it over several posts as I have found that too many pics in one post causes problems.

You should be able to follow but if you have specific question on specific diagrams feel free to ask.

Firstly you should sketch this and realize immediately that ABCD is a kite. You should also already know that the diagonals of a kite bisect each other at right angles.

Diagram 1

Let x be the largest

the one below it   is   x-1  and the other is x-2

added together =1341

x +  x-1 + x-2 =1341

3x-3  = 1341

3x = 1344

x = 448

thank you!!

15 30 45 60 75 90 105 120 135 150 165 180 195 210 225 240

48 96 144 192 240

Lets see

\(n-2(n+1)!+6n!=3(n+1)\\ n-2n!(n+1)+6n!=3n+3\\ -2n!(n+1)+6n!=2n+3\\ n![-2(n+1)+6]=2n+3\\ n![-2n-2+6]=2n+3\\ n![4-2n]=2n+3\\ n!=\frac{2n+3}{-2n+4}\\ \)

if n=1       LHS=1     RHS=5/2 = 2.5

IF N=2      LHS=2     RHS=7/0  UNDEFINED

IF N=3      LHS=6      RHS=9/-2  = -4.5

For all higher integer values of n the right hand side will be negative so there are no solutions to this.

(unless i made some stupid algebraic error)

Did you give it a go already?

\(4 sec^2 x + 2 tan^2 x − 6 = 0\\~\\ hint:\\ sin^2x+cos^2x=1\\ \text { if i divide by }cos^2x \text{ then I get}\\ tan^2x+1=sec^2x \)

Now I did one of these for you just a moment ato so you should be able to finish it.

Thanks guest fort spoting my mistake. I am sorry, I was doing another homework and didn't double check my typing.Yeah , you are right，the correct formula for sin(z) is sin(z)=sin(x+iy)= sin(x)*cosh(y)+i*cos(x)sinh(y). The image of strip R under mapping f(z)=sin(z) is  quadrant+the fourth quadrant+the positive real axis of complex plane. Let w=f(z),then Re(w) is the set positive number and Im(w) is real number.Then the image is an open set,isn't it? Where the definition of open set is  A set is open if all its points are interior points,and the definition of interior point is A point a ∈ G is an interior point of G if some open disk with center a is a subset of G. And open disk is the inside of this circle with center a and radius r; we use the notation D[a,r] := {z ∈ C : |z − a| < r} .

See a similar problem posted on stack exchange

Hi Critical, 

In future could you please put only one question per post.

second question -

82+88+91+83 = 344

(344+F)/6   will give the score out of 100  (percentage)  

So if he scores a B in what range will F have to lie?

\(80\le \frac{344+F}{6}\le 89\)

Anyway, since only on is an outlier it must be the highest or the lowest one!

So if you totally guess you would have 50% chance of getting it right  :)

Also this means that you only need to check one end, it is either the one that you want or else it must be the one at the other end :)

\(12 tan^3 x = 4 tan x\\ let\;\; y=tanx\\ 12y^3 = 4y\\ 12y^3 -4y=0\\ 4y(3y^2 -1)=0\\ 4y(\sqrt{3}y -1)(\sqrt{3}y +1)=0\\ 4y=0,\;\;\;or\;\;\;\sqrt{3}y -1=0\;\;\;or\;\;\; \sqrt{3}y +1=0\\ y=0,\qquad y=\frac{1}{\sqrt{3}}\qquad or\qquad y=\frac{-1}{\sqrt{3}}\\ tanx=0,\qquad tanx=\frac{1}{\sqrt{3}}\qquad or\qquad tanx=\frac{-1}{\sqrt{3}}\\ x=n\pi,\qquad x=n\pi\pm \frac{\pi}{6}\qquad n\in Z\)

With the first on I would just graph it.

Put the time on the horizongal (bottom) axis and temperature on the vertical axis.

Then you can just read the answer of the graph.

Lol I had the wrong idea didn't I lol

Well...what terms do give you  trouble  ???

I think this situation  is fairly difficult to evaluate....but I believe that the answer involves something  known as "Stirling Numbers of the Second Kind"

The  number of ways of distributing  6  balls into 3  indistinguishable  boxes  [  assuming that one or more boxes may be empty ]   is given by this sum :

S2 ( 6,1)   +  S2(6,2)  + S2(6,3)  =

1     +           31        +    90 =

122  ways

P.S.  - If someone knows more about this....corrections are welcome  !!!!!

Yeah.....you probably would have at some point....some of these trig equations can be real "stumpers"  !!!!

Thank you so much.  I would have never got that. 

Good deal CP

We just did one, here   :  https://web2.0calc.com/questions/test_12

Just substitute   "x"   for "N"  and you got it  !!!!!

5cos^3 (x)   =  5 cos (x)      ...... subtract 5 cos (x)   from both sides

5cos^3(x)  - 5 cos (x)    = 0        .....divide through by  5

cos^3(x) - cos (x)  =  0      ....factor

cos  (x )  ( cos^2 (x)  - 1)   = 0

We have two  equations to consider

cos (x )  =  0     and this happens at   pi /2 and 3pi/2    in the given interval

Also

cos^2 ( x)   - 1   = 0       ....add 1 to both sides

cos^2 (x )   =  1      take both roots

cos (x)  = 1            or       cos (x)  = -1

This happens at               This happens at

x = 0                                     x =      pi   

So...the solutions are

x = 0 ,   pi/2,   pi,  3pi/2

Here's graph to confirm this : https://www.desmos.com/calculator/jx5rslbflo

Ok. Well, if the boxes are all the same, then, I would say, the general way to organise it would be, in number order, or, color order, or sets that coordinate the most... 

Doesn't sound too complicated!

Just come up with a realistic equation! Maybe something like.... Ron had x amount of marbles... Etc.

Distinguishable as in each ball is different, instead of all of the balls having a 1 on them, they each have a different number like 1,2,3..... or they are different colors/patterns, ect.

The only one I don't struggle with is mainly the term equilateral.... 

Perfect!

Someone else helped me, the answer is 5.