Suppose a ball is thrown upward to a height of h 0 meters. After each bounce, the ball rebounds to a fraction r of its previous height. Let h Subscript n be the height after the nth bounce. Consider the following values of h 0 and r. Complete parts (a) and (b) below. h0=30, r=0.25 a- find the first 4 terms of the sequence of heights(hn)
b- find a general expression for the nth term of the sequence (hn)

A)1st term:45

2nd term:48.75

3rd term:49.6875

4th term:49.921875

B) Sₙ = h₀ + 2h₀((∞, n=1) Σrⁿ)

Step-by-step explanation:

We are given;

h₀ = Initial height of the ball = 30

r = Rebound fraction = 0.25

a) The arithmetic sequence of bouncing balls is given by the following;

Sₙ=h₀+2h₀(r¹+r²+r³+r⁴rⁿ)

The first term of the sequence is;

S₁ = h₀ + 2h₀r¹

S₁ = 30 + (2 × 30 × 0.25)

S₁ = 45

The second term of the sequence is;

S₂ = h₀ + 2h₀(r¹+r²)

S₂ = 30 + (2 × 30 × (0.25 + 0.25²)) = 48.75

The third term of the sequence is;

S₃ = h₀ + 2h₀(r¹ + r² + r³) = 30 + (2 × 30 × (0.25 + 0.25² + 0.25³)) = 49.6875

S₄ = h₀ + 2h₀(r¹ + r² + r³ + r⁴)

S₄ = 30 + (2 × 30 × (0.25 + 0.25² + 0.25³ + 0.25⁴)) = 49.921875

B) The general expression for the nth term of the sequence is;

Sₙ = h₀ + 2h₀((∞, n=1) Σrⁿ)

because they could drop even lower..

the answer is b. 60. each interior angle in 120, so the exterior angle must be 60 in order to add up to 180

step-by-step explanation: