Find x. (Finding an angle measure given extended triangles) Please help. Thanks!

x = 31º

Step-by-step explanation:

The verticle angle at the bottom left is 33º

The supplementary angle on the bottom right is 180 - 98 = 82

Sum of the angles in a triangle is 180

(34 + x) + 33 + 82 = 180

Combine like terms

x + 149 = 180

Subtract 149 from both sides

x = 31º

x = 31º

The verticle angle at the bottom left is 33º

The supplementary angle on the bottom right is 180 - 98 = 82

Sum of the angles in a triangle is 180

(34 + x) + 33 + 82 = 180

Combine like terms

x + 149 = 180

Subtract 149 from both sides

x = 31º

Step-by-step explanation:

this may not be at all, but i use the same website you're using for school, and if i'm stuck doing one of these i try to look for it on the section.

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.