geno3141

Using the formula:  D  =  R · T

Downhill:  24  =  R · T

Uphill:       24  =  (R- 6) · (T + 9)     

Downhill:  24  =  R · T     --->     T  =  24 / R

Uphill:      24  =  (R- 6) · (T + 9)   

                      =  (R- 6) · (24/R + 9)                 

                      =  24 + 9R - 144/R - 54

                  0  =  9R - 144/R - 54

                  0  =  9R² - 54R - 144

                  0  =  R² - 6R - 16

                  0  =  (R - 8)(R + 2)

Rate =  8 m/s

1)  Find the radius of the inscribed circle by using the formula:  Area = pi·r².

2)  Find the length of each side of the equilateral triangle by using the formula:  

                     radius of the incircle  =  side of the equilateral triangle / 3

3)  Find the area of the equilateral triangle by using the formula:  Area  =  ( sqrt(3) / 4 ) · side of equilateral triangle

If the sides of the triangle are in the ratio of 1:5 and there are two short sides, the ratio of the three

sides of the triangle are 1:1:5.

This will not form a triangle.

Find the area of parallelogram(ABCD) by multiplying its base (CD) by its height (EG).

One form of the Law of Cosines is:  angle(C)  =  cos^-1(   (a² + b² - c²)  /  (2ab)  )

This can be used to find the size of angle(C).

1)  Start with the equation  x + 1/x  =  3

      Multiply both sides by x, giving you a quadratic.

      Solve this quadratic (use the quadratic formula).

2)  Take each of the two answers from part 1) and place them into the expression:  x3 + 1/x3

(WY - VX) / 2  =  U     --->     (WY - VX) / 2  =  30     --->     WY - VX  =  60

360 - [ arc(XY) + arc(VW) ]  =  WY + VX     --->     360 - [ 120  + 130 ]  =  WY + VX

                                                                     --->     110  =  WY + VX

Putting the two equations together:       WY + VX  =  110

                                                               WY - VX  =  60

                                                                       2WY  =  170     --->     WY  =  85     --->     VX  =  25

The formula for the sum of an arithmetic series is:  Sum  =  n(a + l) / 2

where  n = number of terms     a = first term    l = last term

The sum of the first 3n positive integers is:  Sum  =  3n(1 + 3n) / 2

The sum of the first n positive integers is:    Sum  =  n(1 + n) / 2

The equation becomes:  3n(1 + 3n) / 2  =  150  +  n(1 + n) / 2

Solving:                               3n(1 + 3n)  =  300 + n(1 + n)

                                               3n + 9n²  =  300 + n + n²

                                              8n² + 2n - 300  =  0

                                        4n² + n - 150  =  0

                                     (4n + 25)(n - 6)  =  0

                                                           n  =  6

The sum of the first 4n positive integers is:  Sum  =  4n(1 + 4n) / 2

Since 4n = 24, the sum is ...

If  1 - sqrt(3)  is a solution,  1 + sqrt(3)  will also be a solution.

If  2 + sqrt(3)  is a solution,  2 - sqrt(3)  will also be a solution.

The polynomial, in factored form is:  [ x - (1 - sqrt(3)) ] · [ x - (1 + sqrt(3)) ] · [ x - (2 + sqrt(3) ] · [ x - (2 - sqrt(3) ]  =  0

Expand this expression for your answer.

The center of the circumcircle of a right triangle is the midpoint of the hypotenuse.

This makes the radius of the circumcircle one-half the diameter = 58 / 2  =  29.

The area of the circumcircle is  pi · 29²  =  841 · pi

By a well-known theorem (a polite way to say "look it up"), the radius of the incircle of a right triangle is found

by adding the two sides together, subtracting the hypotenuse and then dividing this answer by 2.

The radius of the incircle is (40 + 42 - 58) / 2  =  12

The area of the incircle is  pi · 12²  =  144 · pi

Subtracting, the answer becomes  697·pi