The measure of minor arc JL is 60°. Circle M is shown. Line segments M J and M L are radii. Tangents J K and L K intersect at point K outside of the circle. Arc J L is 60 degrees. What is the measure of angle JKL? 110° 120° 130° 140°

angle JKL =  120 degrees

Step-by-step explanation:

Since arc JL is 60 degrees, the central angle is also 60 degrees.

Point K is the intersection of tanges at J and L, therefore KJM and KLM are 90 degrees.

Consider quadrilateral JKLM whose sum of internal angles = 360.

Therefore

angle JKL + angle KLM + angle LMJ + angle MJK = 360 degrees

angle JKL + 90 + 60 + 90 = 360

angle JKL = 360 - 90 - 60 -90 = 120 degrees

step-by-step explanation:

b and c

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.