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a^3 + b^3 + c^3 = -15.

From center "O" draw a radius to either endpoint of the chord.

This radius will be length "R".   

From center "O" complete the radius to the intersection point of the two straight lines. 

This segment of the radius will be length "R" – 8.  

You have just created a right triangle. 

Use Pythagoras' Theorem                               R²  =  (R – 8)² + 50²    

Multiply it out                                                    R²  =  R² –16R + 64 + 2500     

Subtract R² from both sides                               0  =  –16R + 64 +2500

Add 16R to both sides                                    16R  =  64 + 2500

Add the constants                                           16R  =  2564

Divide both sides by 16                                       R  =  160.25

If you put the line equations in the form  y = mx + b   the slope is m

First one  y -ax =2

                y = ax + 2        the slope of this line is 'a'

                                       perpindicular slope will be  - 1/a

Line 2  x + 23y = 138

                 23y = -x + 138    divide both sides by 23

                     y = -1/23  x  + 6       slope is -1/23     

                         for these lines to be perpindicular:    - 1/a  = -1/23        so a = 23

Here is a graph:

https://www.desmos.com/calculator/fqvxr43leu

Here is a graph if you are stuck.....

https://www.desmos.com/calculator/94brhqodjk

vertex is directly between the focus (2,6) and directrix x=0

   so   vertex is  1,6

this parabola opens to the right.....     can that get you started?

Please explain the method used.

     a  is a multiple of 3     say 9

     b is a multiple of 6     say 12        a - b  = -3    is NOT a multiple of 6    is NOT a multiple of 2

                                                             b IS a multiple of 3     and a-b IS a multiple of 3

A, B   

5  choices for first digit leaves

4 choices for the second digit leaves

3 choices for the third digit....

  5 x 4 x 3 = 60 possibilities

use constructive counting and just have patience and try, that's how i do it.

Let z = a + bi.  Then from the given equations, a = plus or minus 3, and b = plus or minus 5.  So, the quadrialteral is a 6 by 10 rectangle, which has an area of 60.

You should have 20 DISTINCT permutations as follows:

{4, 5, 6, 6, 6}, {4, 6, 5, 6, 6}, {4, 6, 6, 5, 6}, {4, 6, 6, 6, 5}, {5, 4, 6, 6, 6}, {5, 6, 4, 6, 6}, {5, 6, 6, 4, 6}, {5, 6, 6, 6, 4}, {6, 4, 5, 6, 6}, {6, 4, 6, 5, 6}, {6, 4, 6, 6, 5}, {6, 5, 4, 6, 6}, {6, 5, 6, 4, 6}, {6, 5, 6, 6, 4}, {6, 6, 4, 5, 6}, {6, 6, 4, 6, 5}, {6, 6, 5, 4, 6}, {6, 6, 5, 6, 4}, {6, 6, 6, 4, 5}, {6, 6, 6, 5, 4} = 20 permutations.

To find the sequences of rolls, we need to find the ways to arrange the numbers 6,4,6,6,5 where the 6s are indistinguishable. You should be able to follow through.

Use the permutations formula:

5 nPr 3 = 60

5! / (5 - 3)! =120 / 2 = 60 such numbers

Fix one teacher T_1 at the top of the circle. T represents teachers, and 's' students. Find the ways to choose the students and teachers.

\(T_1,S_1,T_2,S_2,T_3,S_3,T_4,S_4\)

All of the statements are true!  So the answer is in fact "A, B, C, D".

Use constructive counting.

Hmm, it's incorrect. Try again.

There are C(5,3) = 10 ways to choose the position of the 3s, then C(5,1) = 5 ways to choose the position of the 4, and C(5,1) = 5 ways to choose the position of the 1, so the number of sequences is 10*5*5 = 250.

  Gold  triangle  

Area         >  A = 20 u² 

Side         >  X  = ?                      X = sqrt [(A/√3)*4]        X = 6.796 

  Red  triangle  

Area         >  B = 45 u²

Side         >   Y = ?                       Y = sqrt [(B/√3)*4]         Y = 10.194 

 Blue (large) triangle  

Area         >  C = ?                       C = (Z/2)² * √3               C = 125 u²  

Side         >   Z = ?                        Z = X + Y                    Z = 16.99     

 Blue (small) triangle 

Side         >  a = 6.796

Side         >  b = 10.194

Side         >  c = ?                        c² = sqrt[a² + b² - 2ab*cos(q)]     c = 8.99  

Angle       >  q = 60°

Semiperimeter > s = ?                       s = (a + b + c)/2             s = 12.99 

Area         >  A_b = ?                      A_b = sqrt[ s(s-a)(s-b)(s-c)]     A_b = 30 u² 

 Yellow triangle 

Area         >  D = ?                      D = C - 3*A_b             D = 35 u²       

Acute scalene triangle.
Sides: a = 17 b = 25 c = 28

Area: T = 210
Perimeter: p = 70
Semiperimeter: s = 35

Angle ∠ A = α = 36.87° = 36°52'12″ = 0.644 rad
Angle ∠ B = β = 61.928° = 61°55'39″ = 1.081 rad
Angle ∠ C = γ = 81.203° = 81°12'9″ = 1.417 rad

 Calculate the heights of the triangle from its area.
There are many ways to find the height of the triangle. The easiest way is from the area and base length. The area of a triangle is half of the product of the length of the base and the height. Every side of the triangle can be a base; there are three bases and three heights (altitudes). Triangle height is the perpendicular line segment from a vertex to a line containing the base.

Height: ha = 24.706
Height: hb = 16.8
Height: hc = 15