ElectricPavlov

The unknown, x, is how many papers he delivered

    money from papers = .10 * x = ,1x

$ 5 + .1x  = 20.60     Solve for x to find the number of papers he delivered  (the Q does not ask for this)

1 in/3ft  = x /12ft

1/3  x 12 = x =  4 inches

y = -x^2 + 5x      max point will be   - b/2a

                                -5/(-2) = 2.5    this is the value of 'x' at the high point (y)

y high point =  - (2.5)^2 +5(2.5) = - 6.25 + 12.5 = 6.25 ft   = 6 ft 3 inches         DUCK ! 

Standard form    y = ax^2 + bx + c

 2x^2−12x−3y+12=0    re-arrange so y term is on one side

3y = 2x^2-12x+12        divide through by 3

y =   2/3 x^2 - 4x + 4          done.

Three angles of a triangle add to 180 degrees

middle angle = m

smallest  = m-10

largest = 2m

add 'em     m-10   +  m  + 2m = 180

                4m -10 = 180

                  4m = 190

                    m = 47.5 degrees

                    smallest = m-10 =   37.5 degrees

                    largest = 2m = 95 degrees

Use something like this

Tan 37 = opp/adj  =   1250/x    where x = horiz distance

.7535 = 1250/x      

x=1658.8 ft

A = x votes

B = 2x votes

C = x-15000 votes     added together = 109000

x   + 2x   +  x - 15000 = 109000

4x - 15000 = 109 000

4x = 124000

x = 31 000        B = 2x = 62000 votes

  Interesting..... I was going to graph them on desmos....just had not done so yet!  

From 3 pounds to 9 pounds    is a gain of 6 pounds

6 pounds/ 1 pound/month = 6 months

Equate them.....because when they are equal is when they have an intersection/common point

3x^2+4y^2+3x−2y+3=3x^2−2y^2+3x−y+3      delete 3x^2   and 3x from both sides

4y^2 -2y+3 = -2y^2-y+3      gather like terms on one side

6y^2 -y = 0

y(6y-1) = 0      y = 0    or 1/6      now that you have values for 'y'   sub back into one of the original equations to fing the corresponding 'x' values.

SO there is 2 points of intersection.