Pascal wants to find the sum of the measures of the interior angles of a polygon that has n sides. how could he do this?

Answer with explanation:

Smallest Polygon = Triangle , which has 3 sides

Sum of angles of triangle = 180°→→, Using the angle sum property of triangle.

We can find the Sum of angles of triangle, using the formula

   = (n-2)×180°

  = (3-2)×180°

 = 1 × 180°

 = 180°

A Quadrilateral has four sides.

Sum of angles of Quadrilateral = 360°

Or , using the formula

=(4-2)×180°

=2 × 180°

=360°

For, n sided Polygon

Sum of it's Interior angles = (n-2) × 180°

To prove this, we should divide that polygon into triangles, and multiply number of triangles by angle of 180°.

For, example, Quadrilateral can be divided into two triangles , sum of interior angles of Quadrilateral =2 × 180°=360°

600,000,000,000, or 6e14

step-by-step explanation:

answer

a) the midrange of the data set shown is 85.5

step by step explanation

let's find the midrange.

the midrange is nothing but add the greatest value and the least value, then divide it by 2.

here the greatest value = 99

the least value = 72

midrange = (99 + 72)/2

= 171/2

midrange = 85.5

you.