A ball was dropped from a height of 60m. Each time it hit the ground, it bounced up 1/2 (half) of the height that it dropped. Ho
1 answer:
Answer:
The ball traveled 116.25 m when it hit the ground for the fifth term
Step-by-step explanation:
This is a geometric progression exercise and what we are asked to look for is the sum of a GP.
The ball was dropped from a height of 60 m. This means that the initial height of the ball is 60 m.
First value, a = 60
Each time it hit the ground, it bounced up 1/2 (half) of the height that it dropped.
This is the common ratio, r = 1/2 = 0.5
The number of terms it hits the ground is the number of terms in the GP.
number of terms, n = 5
The distance traveled by the ball when it hit the ground for the fifth term will be modeled by the equation:
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Answer:
7. cos(X) = 3/5 = 0.6
cos(Y) = 4/5 = 0.8
8. cos(X) = 15/17 ≈ 0.8824
cos(Y) = 8/17 ≈ 0.4706
9. cos(X) = 1/2 = 0.5
cos(Y) = (√3)/2 ≈ 0.8660
Step-by-step explanation:
The trigonometric ratio for cosine is given as follows;
7. The adjacent leg length to ∠X = 27
The length of the hypotenuse side of the triangle = 45
∴ cos(X) = 27/45 = 3/5 = 0.6
Cos(X) = 3/5 = 0.6
The adjacent leg length to ∠Y = 36
∴ cos(Y) = 36/45 = 4/5 = 0.8
cos(Y) = 4/5 = 0.8
8. The adjacent leg length to ∠X = 15
The length of the hypotenuse side of the triangle = 17
∴ cos(X) = 15/17 ≈ 0.8824
The adjacent leg length to ∠Y = 8
∴ cos(Y) = 8/17 ≈ 0.4706
9. The adjacent leg length to ∠X = 13
The length of the hypotenuse side of the triangle = 26
∴ cos(X) = 13/26 = 1/2 = 0.5
Cos(X) = 1/2 = 0.5
The adjacent leg length to ∠Y = 13·√3
∴ cos(Y) = 13·√3/26 = (√3)/2 ≈ 0.8660
cos(Y) = (√3)/2 ≈ 0.8660.
Answer:
x=6, x=0
Step-by-step explanation:
+
= 25
+
= 25
x=6, x=0