Henrietta solved a rational equation. her work is shown. how could she solve it in a simpler way? a: she could use the least common multiple of the denominator. b: she could use the distributive property correctly. c: she could factor the quadratic expression correctly. d: she could combine like terms correctly.

(a) (8%) compute the probability of an even integer among the 100 integers 1! , 2! , 3! , until 100! (here n! is n factorial or n*(n-1)*(n-2) *… 1) (b) (16%) compute the probability of an even integer among the 100 integers: 1, 1+2, 1+2+3, 1+2+3+4, …., 1+2+3+… + 99, and 1+2+3+… + 100

Refer to the table below if needed. second quadrant third quadrant fourth quadrant sin(1800- - cos(180° -) tan(180°-e) =- tane cot(1800-0) 10 it to solo 888 sin(180° +c) = - sine cos(180° +) =- cose tan(180° +c) = tane cot(180° +o) = cote sec(180° + c) = - seco csc(180° +2) = - csce sin(360° -) =- sine cos(360° -) = cose tan(360° - e) =- tane cot(360° -) = -cote sec(360° -) = seco csc(360° -) = csco sec(180° -) = csc(180° -) = csca 1991 given that sine = 3/5 and lies in quadrant ii, find the following value. tane

What is the equation in slope-intercept form of the linear function represented by the table? y -18 -1 -8 4 2 912 o y=-2x-6 o y=-2x+6 o y=2x-6 o y = 2x+6