London and her children went into a grocery store and will buy apples and
1 answer:
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Answer:
The maximum amount of red kryptonite present is 33.27 g after 4.93 hours.
Step-by-step explanation:
dy/dt = y(1/t - k)
separating the variables, we have
dy/y = (1/t - k)dt
dy/y = dt/t - kdt
integrating both sides, we have
∫dy/y = ∫dt/t - ∫kdt
㏑y = ㏑t - kt + C
㏑y - ㏑t = -kt + C
㏑(y/t) = -kt + C
taking exponents of both sides, we have
when t = 1 hour, y = 15 grams. So,
(1)
when t = 3 hours, y = 30 grams. So,
(2)
dividing (2) by (1), we have
taking natural logarithm of both sides, we have
-2k = ㏑(2/3)
-2k = -0.4055
k = -0.4055/-2
k = 0.203
From (1)
Substituting A and k into y, we have
The maximum value of y is obtained when dy/dt = 0
dy/dt = y(1/t - k) = 0
y(1/t - k) = 0
Since y ≠ 0, (1/t - k) = 0.
So, 1/t = k
t = 1/k
So, the maximum value of y is obtained when t = 1/k = 1/0.203 = 4.93 hours
<u>So the maximum amount of red kryptonite present is 33.27 g after 4.93 hours.</u>
Note that 6% converted to a decimal number is 6/100=0.06. Also note that 6% of a certain quantity x is 0.06x.
Here is how much the worker earned each year:
In the year 1985 the worker earned <span>$10,500.
</span>In the year 1986 the worker earned $10,500 + 0.06($10,500). Factorizing $10,500, we can write this sum as:
$10,500(1+0.06).
In the year 1987 the worker earned
$10,500(1+0.06) + 0.06[$10,500(1+0.06)].
Now we can factorize $10,500(1+0.06) and write the earnings as:
$10,500(1+0.06) [1+0.06]=
.
Similarly we can check that in the year 1987 the worker earned
, which makes the pattern clear.
We can count that from the year 1985 to 1987 we had 2+1 salaries, so from 1985 to 2010 there are 2010-1985+1=26 salaries. This means that the total paid salaries are:
.
Factorizing, we have
![=10,500[1+1.06+(1.06)^2+(1.06)^3+...+(1.06)^{26}]=10,500\cdot[1+1.06+(1.06)^2+(1.06)^3+...+(1.06)^{26}]](https://tex.z-dn.net/?f=%3D10%2C500%5B1%2B1.06%2B%281.06%29%5E2%2B%281.06%29%5E3%2B...%2B%281.06%29%5E%7B26%7D%5D%3D10%2C500%5Ccdot%5B1%2B1.06%2B%281.06%29%5E2%2B%281.06%29%5E3%2B...%2B%281.06%29%5E%7B26%7D%5D)
We recognize the sum as the geometric sum with first term 1 and common ratio 1.06, applying the formula
(where a is the first term and r is the common ratio) we have:
.
Finally, multiplying 10,500 by 59.17 we have 621.285 ($).
The answer we found is very close to D. The difference can be explained by the accuracy of the values used in calculation, most important, in calculating
.
Answer: D
Here is how much the worker earned each year:
In the year 1985 the worker earned <span>$10,500.
</span>In the year 1986 the worker earned $10,500 + 0.06($10,500). Factorizing $10,500, we can write this sum as:
$10,500(1+0.06).
In the year 1987 the worker earned
$10,500(1+0.06) + 0.06[$10,500(1+0.06)].
Now we can factorize $10,500(1+0.06) and write the earnings as:
$10,500(1+0.06) [1+0.06]=
Similarly we can check that in the year 1987 the worker earned
We can count that from the year 1985 to 1987 we had 2+1 salaries, so from 1985 to 2010 there are 2010-1985+1=26 salaries. This means that the total paid salaries are:
Factorizing, we have
We recognize the sum as the geometric sum with first term 1 and common ratio 1.06, applying the formula
Finally, multiplying 10,500 by 59.17 we have 621.285 ($).
The answer we found is very close to D. The difference can be explained by the accuracy of the values used in calculation, most important, in calculating
Answer: D