H(t) = −16t^2 + 75t + 25
g(t) = 5 + 5.2t
A)
At 2, h(t) = 111, g(t) = 15.4
At 3, h(t) = 106, g(t) = 20.6
At 4, h(t) = 69, g(t) = 25.8
At 5, h(t) = 0, g(t) = 31
The heights of both functions would have been the closest value to each other after 4 seconds, but before 5 seconds. This is when g(x) is near 30 (26-31), and the only interval that h(t) could be near 30 is between 4 and 5 seconds (as it is decreasing from 69-0).
B) The solution to the two functions is between 4 and 5 seconds, as that is when their height is the same for both g(t) and h(t). Actually the height is at 4.63 seconds, their heights are both
What this actually means is that this time and height is when the balls could collide; or they would have hit each other, given the same 3-dimensional (z-axis) coordinate in reality.
Answer:
x = 2
Step-by-step explanation:
Taking antilogs, you have ...
2³ × 8 = (4x)²
64 = 16x²
x = √(64/16) = √4
x = 2 . . . . . . . . (the negative square root is not a solution)
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You can also work more directly with the logs, if you like.
3·ln(2) +ln(2³) = 2ln(2²x) . . . . . . . . . . . write 4 and 8 as powers of 2
3·ln(2) +3·ln(2) = 2(2·ln(2) +ln(x)) . . . . use rules of logs to move exponents
6·ln(2) = 4·ln(2) +2·ln(x) . . . . . . . . . . . . simplify
2·ln(2) = 2·ln(x) . . . . . . . . . . . subtract 4ln(2)
ln(2) = ln(x) . . . . . . . . . . . . . . divide by 2
2 = x . . . . . . . . . . . . . . . . . . . take the antilogs
Answer:
1.55%
Step-by-step explanation:
Fill in the values and solve for the rate using the future value formula.
FV = P(1 +r/n)^(nt)
4720 = 4300(1 +r/12)^(12·6)
(4720/4300)^(1/72) = 1 +r/12
r = 12(4720/4300)^(1/72) -1) ≈ 0.015542
The annual interest rate is about 1.55%.
Answer:
the correct answer is Y=-x-3
Answer:
19 Cds left.
Step-by-step explanation:
The question said he used 31 Cds, so 50 minus 31 is 19.
