6 points are placed on the line a , 4 points are placed on the line b . How many triangles is it possible to form such that their vertices will be the given points, if a is parallel to b ?

The number of possible triangles that can be formed is 132 triangles

Step-by-step explanation:

Given that there are 6 points in line a and 4 points on line b

Number of points on line a = 6

Number of points on line b = 4

To form a triangle, firstly we select 2 points from line a and one point from line b as follows;

Number of ways to select 2 points from 6 = ₆C₂ = 15 ways

Number of ways to select 1 points from 4 = ₄C₁ = 4 ways

The number of triangles = 15×4 = 60 triangles

Next we form triangles by selecting 2 points from line b and one point from line a as follows;

Number of ways to select 2 points from 4 = ₄C₂ = 12 ways

Number of ways to select 1 points from 6 = ₆C₁ = 6 ways

The number of triangles = 12×6 = 72 triangles

The total number of triangles = 72 + 60 = 132 triangles.

given the equation is 8+9z=-6, the first step would be to subtract the 8 from both sides:

8+9z-8=-6-8

since 8 minus 8 = 0, this would cancel out the 8. and since -6 minus 8 equals -14, the equation would look like this

9z = -14