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A regular dodecahedron has 12 congruent sides.

If the center of the circle is C, draw all the radii from C to each of the points P1, P2, ... P12.

Each of the triangles formed will be congruent to each other.

Consider the triangle, triangle(CP1P2).

Side CP1 = 1 and side CP2 = 1.

The central angle angle(P1CP2) = 30o.    (360o / 12o  =  30o)

Use the Law of Cosines on this triangle to find side P1P2:

     (P1P2)2  =  (CP1)2 + (CP2)2 - 2·(CP1)·(CP2)·cos( angle(P1CP2) )

     (P1P2)2  =  12 + 12 - 2·1·1·sqrt(3)/2

This makes the sum work out to 252.

An equilateral triangle of side length 2 units is inscribed in a circle. Find the length of a chord of this circle which passes through the midpoints of two sides of the triangle.

BC = 2       BQ = 1       ∠B = 60º

Height       AQ = tan(B) * BQ = 1.732050808

AO = 2/3 * AQ = 1.154700537              AO = XO

AP = AQ / 2 = 0.866025402

PO = AO - AP = 0.288675134

XP = sqrt( XO² - PO² ) = 1.118033987

XY = 2 * XP = 2.236067975     

Same as geno.

14+1=15, 12+3=15

honestly, i have no idea why it was flagged cause it looks perfectly fine to me

this has happened to me too, and i usually just message Melody to ask why it was flagged

Question 1:

-3 and the bottom left have to multiply to -6, so the bottom left circle is 2. Then 2 and the bottom right number must multiply to 12 so the bottom right circle is 6. Then the middle right square must equal to the product of -3 and 6, which is 18.  So the missing numbers from clockwise are 18, 6, and 2.

Question 2:

Let n be a number for each of these parts.

a)

-4 * n =320

n=320/-4

n= -80

b)

n * 3 =-180

n=-180/3

n= -60

c)

n / -7 = -14

n=-14 * -7

n= 98

d) 

n / -9 = 6

n= 6 * -9

n= -54

Notice that multiplying the denominators x-2 and x+1 give us \(x^2-x-2\).

\(\frac{x+7}{x^2-x-2}=\frac{A}{x-2}+\frac{B}{x+1}\\ \frac{x+7}{x^2-x-2}=\frac{A(x+1)}{x^2-x-2}+\frac{B(x-2)}{x+1}\\ x+7=A(x+1)+B(x-2)\\ x+7=Ax+A+Bx-2B\\ \)

Now look here, the x terms Ax+Bx must result in x(with a coefficient of 1) so A+B=1 and A-2B must result in 7.

So we have a system of equations.

\(A+B=1\\ A-2B=7\\ \text{then:}\\ 3B=-6\\ B=-2, A=3\)

Therefore the pair is \((3, -2)\)

My guess is 15.

Notice that \((x+y)^3=x^3+y^3+3xy(x+y)\). Then we plug in 3 for the (x+y)'s and xy's.

\((x+y)^3=x^3+y^3=3xy(x+y)\\ 3^3=x^3+y^3=3(3)(3)\\ 27=x^3+y^3=27\\ x^3+y^3=0\)

Since  x + y  =  3     --->     y  =  3 - x

Since  xy  =  3     --->     y  =  3/x

Combining these equations:  3/x  =  3 - x     --->     3  =  3x - x²     --->     x² - 3x + 3  =  0

Use the quadratic formula to get:  x  =  ( 3 + i·sqrt(3) ) / 2  and  x  =  ( 3 - i·sqrt(3) ) / 2

[If x is one of the answers, then y is the other.]

Now, cube the answers to get the final answer ...  (it's an interesting answer) ...

The area of the circle is pi·radius²  =  100 pi

If you were to connect the center of the circle to each of the vertices of the interior hexagon, you would

get 6 congruent equilateral triangle and each of these triangles would be congruent to the six 

equilateral triangles "at the points".

A radius to one of the points on the circle = 10, so the height of each of the equilateral triangles is 5.

This means that the base of the triangle is ( 5 / sqrt(3) ) · 2  =  10/sqrt(3)

The area of each triangle is  ½ · ( 10/sqrt(3) ) · 5  =  25 / sqrt(3).

Since there are 12 congruent triangles, the total area is 300/sqrt(3)  =  100·sqrt(3)

The area inside the circle but outside the hexagram =  100 pi - 100·sqrt(3)

Of this area, one-third is shaded  =  ( 100 pi - 100 sqrt(3) ) / 3  

When you are finding the fourth root of ( 4 / 20.25 ) --

the fourth root is the square root of the square root , so:

  fourth root of ( 4 / 20.25 )  =  sqrt( sqrt( 4/20.25) ) =  sqrt( 4/9 )  =  2/3

     because the sqrt( 4/20.25 )  =  2/4.5  or  4/9

No problems with the rest of the problem ...

Check the. Fret 4/20.25.      Is that not 2/3 ?

Hi there! As I was looking to see if this answer was asked before I found it! hope fully these two can give you the correct answer and help you understand!

https://web2.0calc.com/questions/due-today-if-u-want-to-just-give-me-a-hint

https://web2.0calc.com/questions/lots-of-questions 

~Wolf 

ty! this rlly helped!

sumfor(n, 0, 1000, (fibon(n) / 4^n) = 4 / 11

I believe that they can be seated as:

M-F-M-F-M-F-M-F

M-F-M-F-M-F-F-M

M-F-M-F-F-M-F-M

M-F-F-M-F-M-F-M

F-M-F-M-F-M-F-M

In each case there will be 4 x 4 x 3 x 3 x 2 x 1 x 1 = 576 was.

5 x 576 = 2880 ways

If the volume is 389344 and the base is a 92 x 92 square, then the 

height = 389344 / (92²)  =  46.

The area is:  4 x (92 x 46)  + (92 x 92)  =  25392 cm²

Since you want distinct sums, order is not important.

2 + 2 + 2  =  6

2 + 2 + 4  =  8

2 + 2 + 8  =  12

2 + 4 + 4  =  10

2 + 4 + 8  =  14

2 + 8 + 8  =  18

4 + 4 + 4  =  12

4 + 4 + 8  =  16

4 + 8 + 8  =  20

8 + 8 + 8  =  24

Sum  =  140

Let the common ratio be r.

AB = x

BC = x·r

AC = x·r²

AC / AB  =  x·r² / x  =  r²  

To find r: the Pythagorean Theorem:  ( x )² + ( x·r )²  =  ( x·r² )² 

                                                                   x² + x²·r²  =  x²·r⁴ 

                                                                                0  =  x²·r⁴ - x²·r² - x² 

                                                                                0  =  x²·( r⁴ - r² - 1 )

  Divide both sides by x²:                                        0  =  r⁴ - r² - 1

  Using the quadratic formula:                                r²  =  [ 1 + sqrt( 1 + 4 ) ] / 2

                                                                               r²  =  ( 1 + sqrt(5) ) / 2

AC / AB  =  r²  =  ( 1 + sqrt(5) ) / 2              (phi)

Since OQ = OR, angle(OQR) = angle(ORQ)

angle(O) + angle(OQR) + angle(ORQ)  =  180.

angle(O) =  180 - angle(OQR) - angle(ORQ)

angle(O) = 180 - 2·angle(ORQ)

angle(O) = arc(QR)

angle(QPR) = ½·arc(QR) = ¼·angle(O)   --->    2·angle(QPR)  =  angle(O)

--->                                2·angle(QPR)  =  180 - 2·angle(ORQ)

         2·angle(QPR)  +  2·angle(ORQ)  =  180

                angle(QPR)  +  angle(ORQ)  =  90

In triangle(ABC), angle(B) = angle(C) = 45^o.

In triangle(AFP), angle(AFP) = angle(APF) = 45^o.

Let  AP = x   --->   PF = x·sqrt(2).

Since AC = 6 and AP = x   --->   PC = 6 - x   --->   DP = (6 - x) / sqrt(2).

Since FSDP is a square   --->   PF = DP   --->   x · sqrt(2)  =  (6 - x) / sqrt(2).

                                                                  --->            x · 2  =  6 - x

                                                                  --->               2x  =  6 - x

                                                                  --->               3x  =  6

                                                                  --->                 x  =  2

Since  PF = x·sqrt(2)   --->   PF  =  2·sqrt(2)

From here, you can calculate the area of the square.

Thanks you are life savers i have test coming up

You will get two answers when you apply the quadratic formula; these will be two x-values.

Take the first x-value and place it into either of the two original equations to get a y-value;

    this x-value and y-value will be one of the points of intersection.

Then, take the other x-value and place it into either of the two original equations,

   this x-value and y-value will be the other point of intersection.

                  y  =  4x² - 28x - 17

          y + 17  =  4x² - 28x                    (add 17 to both sides)

           y + 17  =  4(x² - 7x)                   (factor)

   y + 17 + 49  =  4(x² - 7x + 49/4)        (complete the square)

           y + 66  =  4(x - 7/2)²                  (factor)

Now, select the vertex from the equation.