After being dropped a certain ball always bounces back to 2/5 of the height of its previous bounceAfter the first bounce it reac
1 answer:
Answer:
8 inches.
Step-by-step explanation:
In the first bounce, the height that the ball reaches was 125 inches.
If the ball always bounces back to of the height, then after 2nd bounce it will reach
inches height.
Again, after 3rd bounce, it will reach inches height.
And, after 4th bounce, it will reach inches height. (Answer)
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Answer:
Answer of 17 is ㏒(+15x), Answer of 33 is x = 8 , Answer of 35 is x = ㏒10/㏒2 , Answer of 37 is x = -㏒12/㏒8 and Answer of 39 is x = 5
Step-by-step explanation:
17. ㏒x + ㏒(x+15)
Using property ㏒a + ㏒b = ㏒a×b
∴ ㏒x + ㏒(x+15)
㏒x×(x+15)
㏒(+15x)
The answer is ㏒(+15x)
33. 2^(x-5) = 8
2^(x-5) = 2^3
Using property 2^a = 2^b
Then a = b
∴x-5 = 3
x = 8
The answer is x = 8
35. 2^x = 10
Taking log on both sides gives
㏒2^x = ㏒10
x×㏒2 = ㏒10
x = ㏒10/㏒2
The answer is x = ㏒10/㏒2
37. 8^-x = 12
Taking log on both sides gives
㏒8^-x = ㏒12
-x×㏒8 = ㏒12
x = -㏒12/㏒8
The answer is x = -㏒12/㏒8
39. 5(2^3 × x) = 8
5(8×x) = 8
x = 5
The answer is x=5
I’ve attached my work...
Hope it helps! Have a nice day :)
Hope it helps! Have a nice day :)