4 When a ball bounces, the heights of consecutive bounces form a geometric sequence. The height of the first bounce is

121 centimeters and the height of the third bounce is

64 centimeters. To the nearest centimeter, what is the height of the

fifth bounce?

34 cm

Step-by-step explanation:

The nth term of a geometric progression Tₙ = arⁿ⁻¹ where a = first term, and r = common ratio.

Given that our first term, T₁ = ar¹⁻¹ = a = 121 cm and our third term T₃ = ar³⁻¹ = ar² = 64 cm

Then, T₃/T₁ = ar²/a = r²

r² = T₃/T₁

r² = 64 cm/121 cm

r = √64/121

r = 8/11

So, the height of our fifth bounce T₅ = ar⁵⁻¹

= ar⁴

= 121 cm × (8/11)⁴

= 121 × 4096/14641

= 4096/121 cm

= 33.9 cm

≅ 34 cm to the nearest cm

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this

32/80  

91/140

the last one 54 doesn't go into 45. 54/100 ?

step-by-step explanation: