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The side  of the square   =  3√2

So  the  legs  of the small white triangle at the top   = (1/2)  of this = (3/2)√2

So....the area  of  one of the  small  white  triangle =  (1/2) [(3/2)√2]^2 =  9/4  units^2

So.....the area of the 4 small white triangles=  4 (9/4)  =  9 units^2

And the area of the circle =  pi (3)^2 =  9pi  units^2

So......the shaded area  =

[ 9pi  - 9]  units^2    = 

9 [ pi - 1 ]  units^2   ≈  19.27  units^2  

We can let the eight points be

A = (1,0)   B  = (1/√2 , 1/√2) C  =  (0, 1)  D  = (-1/√2, 1/√2)  E  = (0, -1)

F  = (-1/√2, -1/√2)   G  =  (0, -1)    H  = (1/√2, -1/√2)

Let A  be the selected  point  that we will use to calculate the distances to every other point

We can use the distance formula to calculate the distance from A to the other seven points [which you should be able to work with  ]

I will  use  WolframAlpha  to calculate these

AB  =  √ [ 2 -√2]

AC = √2

AD  = √[2 + √2 ]

AE  =  2

AF  = √[2 + √2 ]

AG = √2

AH  = √[2 - √2]

And computing these  products we have

2 * √2 * √2  * (√[2 - √2])^2  * ( √[2 + √2] )^2  =

2 * 2   *  ( 2 - √2) ( 2 + √2)  =

4  *  [ 4  - 2  ]  =

4 *  2   =

8

Note that triangle  RQP   is similar to triangle  TSP

Therefore

 QP / RP  = SP / TP

{PT +QT ]  / QP  = PS/ TP

[11 + 10] / 35  =  PS /  11

21 / 35 =  PS / 11

21 * 11/ 35  = PS

(21/35)* 11   =PS

(3/5) * 11  = PS

33/5  =  PS  =  6.6

Let the meaure  of  the smaller angle   = A

And the measure of the  other known angle  be  A+ 60

And by the Law  of Sines

sin (A + 60)               2          

_________  =         ___         so we have

  sin A                        1

sinAcos60  + sin60cosA

___________________     =      2

           sin A

sinA  + √3cosA

_____________       =   2

          2sin A

sinA              (√3/2) cot A   =   2

______  +  

2sinA

(1/2)  +  (√3/2)cot A  =  2

(√3/2)cotA  = 3/2

cot A  = 3 / √3

cot A  =   √3

arccot(√3)  = A  =    30°

So  the second known angle  =  A + 60   =  90°.....and this must be the measure of the largest angle in the triangle

that... is a more complicated way. BUT AWESOME! 

Let's make the solution super-formal!  We can write -4 in exponential notation as 4e^(pi*i), so the equation is z^4 = 4e^(pi*i).

By Hamilton's Theorem, the solutions are z = 4^{1/4}*e^(pi*i/4), 4^{1/4}*e^(pi*i/4 + pi/4), 4^{1/4}*e^(pi*i/4 + 2*pi/4), and 4^{1/4}*e^(pi*i/4 + 3*pi/4).  Since 4^{1/4} = sqrt(2) and e^(pi*i/4) = (1 + i)/sqrt(2), the first solution is 1 + i.  Then the other roots work out as

4^{1/4}*e^(pi*i/4 + pi/4) = 1 - i,

4^{1/4}*e^(pi*i/4 + 2*pi/4) = -1 - i, and

4^{1/4}*e^(pi*i/4 + 3*pi/4) = -1 + i.

Why would anyone go to an accounting firm called "Assured Bats"?

The angles of a quadrilateral add up to 360 degrees.  If you add up the angles, and solve the equation, you get q = 25.  Therefore, the largest angle is 5q + 15 = 140 degrees.

Set A =[2, 3, 20, 21] =2,520

Set B =[12, 14, 15 ] = 2,520 

Since \(f(x+y)=f(x)+f(y)\), we can arrive at the following facts:

\(f(10)=f(2+8)=f(2)+f(8)\),

\(f(8)=f(2+6)=f(2)+f(6)\),

\(f(6)=f(2+4)=f(2)+f(4)\), and

\(f(4)=f(2+2)=f(2)+f(2)\).

Using these facts we can write

\(f(10)=f(2)+f(2)+f(2)+f(2)+f(2)\). In addition, since \(f(1)=3\), we can write

\(f(2)=f(1+1)=f(1)+f(1)=3+3=6 \). So

\(f(10)=6+6+6+6+6=30\).

Note that the four triangles EGO, EIO, IHO, and HGO can be rotated, along with their shading, about the line segments EG, EI, IH, and HG respectively to transform FIG. 1 into FIG. 2. So the blue regions in the two figures have equal area. The area of the blue regions in FIG. 2, however, can be calculated easily. Each side of the square has the same length as the radius of the circle, so the shaded (blue) region has area equal to

\(\pi r^2-r^2\) \(=9\pi-9\) square units.

oh wow

Ok, I unfortunately do not have a venn diagram to upload, but I do have one on paper. 

1. Draw a venn diagram with 3 circles that are overlapping, and label the circles their own colors.

2. put 16 in the middle, where they all intersect.

3. Fill in each intersection, but subtract 16 everytime. (for example, 32 have blue and green. so 32-16=16. So put 16 in the blue and green intersection)

4. once you're done. add up the whole circle, including the middle and the intersections. (e.x. 23+16+16=55. and then you subtract 55 from 62, since it says that they have 62 blue ribbons, so in total, you have 7 dogs with just blue ribbons)

5. since the question is asking you for how many dogs dont have any ribbons, you can count all of them up, and subtract that by 100.

(it's 12)

z^4 = -4

z^2 = +- sqrt(-4) = +-2i

z = +- sqrt(+-2i)

=  -1+i  ,   -1-i    ,  1-i    ,  1+i 

actually... it's 3^4... which btw is 81

y + 2x = -19

y = -2x - 19          this is DIRECT variation    k = -2      

(EXAMPLE of INdirect variation would be something of THIS nature  y = -2/x - 19  )

Yes, it will fit diagonally....

3¹ x 3¹ x 3¹ x 3¹ = 3^(1+1+1+1)= 3⁴