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Equilateral triangles, so both angles near C are 60 degrees

   so  angle ACD = 80 + 60 = 140 degrees

       AC and CD are equal in length, so triangle ACD is isoceles

          unlabelled angle A and D are equal  180 - 140 = 40 dgrees  so A and D are 20 degrees each

             BAD = 60 - 20 = 40 degrees  

Yes. I got this problem from my friend. When I searched it up, I found a very similar AMC 10 problem. That's probably where he got it from.

https://artofproblemsolving.com/wiki/index.php/2010_AMC_10A_Problems/Problem_24

https://web2.0calc.com/questions/help_7838

Perimeter = L  +  L + W + W = 2L + 2W

L= 5W + 7  (given)

  or   (L-7)/5 = W

2L + 2W = 86                             (substitute in W = (l-7)/5)

2L + 2 ( L-7)/5   = 86                  (multiply both sides by 5)

10L + 2L - 14 = 430                     (simplify)

12L = 444                                    (divide both sides by 12)

L = 37                                W = 6

Here you go!

https://web2.0calc.com/questions/help_4754
 

angle a + angle b + angle c = 180 because the 3 interior angles of a triangle add to 180.

now, angle a is 40 degrees, because the 40 degree angle and angle a are corresponding angles (parallel lines and transversals).

The triangle is isosceles, so angle a = angle c. Angle c is 40 degrees, and by corresponding angles, the supplement of angle e is 40 degrees.

180-40=140

angle e is 140 degrees

180+140=320

The answer is 320. Sorry if it isn't really well explained, I'm not the best at explaining things.

Edit: 5 seconds after posting, i realize i said 220 instead of 320. Simple math mistake! 

:9

From the cyclic condition, the largest angle of PQRS works out to 150 degrees.

5 spikers, 4 setters, 1 either.      2 person team.

if the odd one is the spiker then  4 pairs

if the odd one is as setter then 5 pairs

if the odd on is not included then 5*4=20 pairs

so that makes   29 possible pairs.

A yellow circle circumscribes a green regular hexagon and is inscribed in a red regular hexagon.

What is the ratio of the areas of the green hexagon to the red hexagon?

Nothing wrong with your strategy CalculatorUser; though a graphical approach might help your guess and check part:

It is (maybe) easiest to start with a sketch.

\(y=-x^2-3x-2\\ y=-(x^2+3x+2)\\ y=-(x+2)(x+1)\)

This is a concave down parabola, roots are -2 and -1.   

Vertex is at x= -3/2

When x=-3/2   y=-9/4 +9/2 -2 = 0.25

Vertex is (-1.5, 0.25)

A quick look at the graph reveals that  y=|x|   is always above the parabola.

Also it is easy to see that the shortest distance betweem them will definitely be somewhere in the domain \( -1.5 \le x\le 0\)

so

the distance betweem them is  

\(Q=|x|-(-x^2-3x-2)\\ Q=|x|+x^2+3x+2\\ \)

But x will be negative so

\(Q=-x+x^2+3x+2\\ Q=x^2+2x+2\\ \)

Q is a concave up parabola so any stat point will be a minimum.

\(\frac{dQ}{dx}=2x+2\)

This equals 0 when x=-1

When x=-1   Q=1-2+2 = 1

So the smallest vertical distance between the graphs is 1 and it occurs when  x=-1

Here is the graph:

Coding:

Q=|x|-(-x^2-3x-2)\\
Q=|x|+x^2+3x+2\\

Q=-x+x^2+3x+2\\
Q=x^2+2x+2\\

Large circle radius   r = 2          Area of a large circle:     A = 2² * pai = 12.566 u²

Small circle radius   r = 1          Area of 4 small circles:   A = pai * 4   =  12.566 u²

This proves that the blue and yellow areas are equal. 

Especially if you are good at maths, live to be 1000000000 years old and have nothing else ot do with your time.

(cos (3x-x) ) = cos 3pi/5

cos 2x = cos 3pi/5

2x = 3 pi/5

x = 3pi/10

This is expected value am I right?

Try these links:

Probably. 

In that case, I would assume trying to solve it guest's way would still work, assuming again that you don't make a math error AND you don't have a time limit...

Circle  radius                r = 2

Angle                            q = 30°

Red hxgn 1/2side         S₁ = ?                S₁ = tan(30) * r         S₁ = 1.1547         

Green hxgn 1/2side      S₂ =?                 S₂ = sin(30) * r         S₂ = 1.000 

Red hxgn area              A₁ =?                 A₁ = ( S₁*r) *6          A₁ = 13.8564 u² 

Green hxgn area           A₂ =?             A₂ =[S₂*sqrt(3)]*6        A₂ = 10.3923 u² 

Ratio                                                                 A₂ : A₁  =   3 : 4              

My strategy:

After analyzing the two equations, we can conclude that A and B must be perfect square integers.

As both of their square roots end up forming integers after being added by eachother.

So I would guess and check from there.

They both should be less than 121 because the largest integer in the equations is 11.

So I would guess all perfect squares from 1 to 121.

Not the best strategy...

Interesting geometry question.

Great answer from Omi67      (Dec19)

Was this question meant to be done without a calculator?

I answered this sometime in the last week.

Go look for my answer.