What is the area of this triangle in the coordinate plane? 5 units² 6 units² 7 units² 12 units² A graph with a triangle drawn between coordinates begin ordered pair 1 comma 1 end ordered pair begin ordered pair 1 comma 4 end ordered pair and coordinates begin ordered pair 5 comma 1 end ordered pair

6 units²

Step-by-step explanation:

The coordinates (1,1) (1,4) (5,1)  form a right triangle as shown in the graph below

The base is  is (1,1) and (5,1) so the length is 5-1 or 4 units

The height is (1,1) and (1,4) so the height is 4-1 or 3 units

Area of a triangle is given by

A = 1/2 bh

    = 1/2 (4*3)

    = 1/2 *12

    = 6 units^2

The variables should add up to equal 1

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.