In the figure, anglerqs is-congruent-to angleqlk. 3 lines are shown. lines s p and k n are parallel. line r m intersects line s p at point q, and intersects line k n at point l. angle r q s is x degrees. angle k l m is (x minus 36) degrees. what is the value of x?

x=108

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

m∠RQS≅m∠QLK -----> by corresponding angles

m∠KLM+m∠QLK=180° -----> by supplementary angles (consecutive interior angles)

we have that

m∠RQS=x° ----> given problem

so

m∠QLX=x°

m∠KLM=(x-36)° ----> given problem

substitute

Dang i wish i could , good luck (i’m bad at math)

Answer: f(x) = -2x^3 +3x^2 +11x -6see attachedan infinite number. since the magnitude of the leading coefficient is not specified, it may be any negative number. (we have chosen the smallest magnitude integer that makes all coefficients be integers.)step-by-step explanation:

1. when "a" is a root of a polynomial, (x -a) is a factor of it. for the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).

in order to make the leading coefficient be negative, we need to multiply this product by a negative number. any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.

then

f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)

= -(2x² +3x -2)(x -3)

= -(2x³ -3x² -11x +6)

f(x) = -2x³ +3x² +11x -6

2. a graph created by the desmos on-line graphing calculator is shown, and the zeros are highlighted.

3. as indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. you want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.