Mrs. williamson had 18 students in her homeroom at the beginning of the school year and 24 students at the end of the year what was the approximate percent increase in the number of students in miss williamson‘s homeroom

There was an increase of 33.33% in Miss Williamson‘s homeroom

Step-by-step explanation:

Percentage increase in the number of students=((Final number-Initial number)/Initial Number)×100

where;

Initial number of students=18 students

Final number of students=24 students

Replacing;

((24-18)/18)×100=(6/18)×100=33.33% increase

There was an increase of 33.33% in Miss Williamson‘s homeroom

The length of the track is approximately 51.7 ft

The track has three sides of the square and the distance round a half of a complete circle

Step-by-step explanation:

The given track shape and measurements are;

The shape on the left side of the track  = Square

The shape on the right side of the track  = Half circle

The area of the square on the the left side of the track  = 128 square feet

Therefore, from the area, A, of a square of side length, s, which is s × s, and letting the side length of the square = s, we have;

Area of the square portion of the track = s × s = s² = 128 ft²

Therefore, s = √(128 ft²) = 8·√(2) ft.

Whereby the side length of the square is bounded by the diameter of the half circle, we have;

Length of the diameter of the half circle = s = 8·√(2) ft.

The length of the perimeter of the half circle = π·D/2 = π × 8·√(2)/2 = π × 4·√(2) ≈ 17.77 ft.

The perimeter of the track, which is the length of the track is made up of the three sides of the square opposite to the half circle and the circumference of the half circle.

Therefore;

The length of the track = 3 × 8·√(2) ft + π × 4·√(2) ft. = 4·√2×(π+6) ≈ 51.7 ft

The length of the track ≈ 51.7 ft

Which gives;

The track has three sides of the square and the distance round a half of a complete circle.