What is the probability that a point chosen at random in the given figure will be inside the larger triangle and outside the smaller triangle? Enter your answer, as a fraction in simplest form, in the box.

P(inside larger triangle and outside smaller triangle) =

(x3 + 6x2 + 2x - 6) • (x - 1)

step-by-step explanation:

polynomial long division :

3.2     polynomial long division

dividing :   x4+5x3-4x2-8x+6

                              ("dividend")

by         :     x-1     ("divisor")

dividend     x4   +   5x3   -   4x2   -   8x   +   6

- divisor   * x3     x4   -   x3            

remainder         6x3   -   4x2   -   8x   +   6

- divisor   * 6x2         6x3   -   6x2        

remainder             2x2   -   8x   +   6

- divisor   * 2x1             2x2   -   2x    

remainder               -   6x   +   6

- divisor   * -6x0               -   6x   +   6

remainder                     0

quotient :   x3+6x2+2x-6   remainder:   0

part a

x     y

0     8

1     7.5

2     7

3     6.5

4     6

5     5.5  

for every hour i deducted .5 inches from the length of the candle

part b

it is a function because for every x value there is only one y value

part   c

for every x value there is only one y value

x       y

0       8

1       7.6

2       7.2

3       6.8

step-by-step explanation:

d. it is a convex pentagon because it has five sides and none of the sides would extend into the inside of the polygon.

step-by-step explanation: