To calculate elapsed time:
Count on in minutes from the earlier time to the nearest hour.
Count on in hours to the hour nearest to the later time.
Count in minutes to reach the later time.
The gap from 9:25 PM to 10 PM is 35 min. The gap from 6 AM to 6:15 AM is 15 min. So far, that accounts for 35+15 = 50 min. The gap from 10 PM to midnight is 2 hours (10+2 = 12). Then the gap from midnight to 6 AM is 6 hours. So we have 2+6 = 8 hours. Add on the minute portion we computed to get 8 hours, 50 minutes.
Answer: 8 hours, 50 minutes
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Answers:
- a) 15000 represents the starting amount
- b) The decay rate is 16%, which means the car loses 16% of its value each year.
- c) x is the number of years
- d) f(x) is the value of the car after x years have gone by
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Explanation:
We have the function f(x) = 15000(0.84)^x. If we plug in x = 0, then we get,
f(x) = 15000(0.84)^x
f(0) = 15000(0.84)^0
f(0) = 15000(1)
f(0) = 15000
In the third step, I used the idea that any nonzero value to the power of 0 is always 1. The rule is x^0 = 1 for any nonzero x.
So that's how we get the initial value of the car. The car started off at $15,000.
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The growth or decay rate depends entirely on the base of the exponential, which is 0.84; compare it to 1+r and we see that 1+r = 0.84 solves to r = -0.16 which converts to -16%. The negative indicates the value is going down each year. So we have 16% decay or the value is going down 16% per year.
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The value of x is the number of years. In the first section, x = 0 represented year 0 or the starting year. If x = 1, then one full year has passed by. For x = 2, we have two full years pass by, and so on.
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The value of f(x) is the value of the car after x years have gone by. We found that f(x) = 15000 when x = 0. In other words, at the start the car is worth $15,000. Plugging in other x values leads to other f(x) values. For example, if x = 2, then you should find that f(x) = 10584. This means the car is worth $10,584 after two years.
These could be be some of the answers to the problem. 6:3, 12:6, and 36:18
12.T
13.F
14.F
15.F.
16.???(where is 16?:O)
17.T
18.T
19.T
20.???(where is 20?)
21.5
22.5
The position function of a particle is given by:
The velocity function is the derivative of the position:
The particle will be at rest when the velocity is 0, thus we solve the equation:
The coefficients of this equation are: a = 2, b = -9, c = -18
Solve by using the formula:
![t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B-b%5Cpm%5Csqrt%5B%5D%7Bb%5E2-4ac%7D%7D%7B2a%7D)
Substituting:
![\begin{gathered} t=\frac{9\pm\sqrt[]{81-4(2)(-18)}}{2(2)} \\ t=\frac{9\pm\sqrt[]{81+144}}{4} \\ t=\frac{9\pm\sqrt[]{225}}{4} \\ t=\frac{9\pm15}{4} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20t%3D%5Cfrac%7B9%5Cpm%5Csqrt%5B%5D%7B81-4%282%29%28-18%29%7D%7D%7B2%282%29%7D%20%5C%5C%20t%3D%5Cfrac%7B9%5Cpm%5Csqrt%5B%5D%7B81%2B144%7D%7D%7B4%7D%20%5C%5C%20t%3D%5Cfrac%7B9%5Cpm%5Csqrt%5B%5D%7B225%7D%7D%7B4%7D%20%5C%5C%20t%3D%5Cfrac%7B9%5Cpm15%7D%7B4%7D%20%5Cend%7Bgathered%7D)
We have two possible answers:
We only accept the positive answer because the time cannot be negative.
Now calculate the position for t = 6:
