Answer:
There will be $4450 left at the end of the year.
Step-by-step explanation:
We first take 11% and multiply it by $5,000. We get 550. This means that the account will lose $550. Next, we take our original amount, $5,000, and subtract $550 from it. We will get $4450.
Multiple primary malignant neoplasms (MPMN) is defined as the presence of two or more malignant tumors of different histological entities in an individual. MPMN are a common occurrence. The incidence among all malignancies ranges from 0.4 to 21% from different studies and countries.
Answer:
A) 26
B) 53
C) 57
Step-by-step explanation:
Your trying to find what the missing letters equal (A, B, C), so looking at the problem there are two "total" rows so first you need to figure out the missing numbers. You don't need to worry about titles but only the "Total", the question doesn't ask to solve for any of those. So when solving depending on what number your going with so lets start with A.
This is a two way table so there is 21, 47 and 17, 43 on each side of A. So we need to solve starting with 21, 47.
T = 47
BV (Boy votes) = 21
Subtract.
47 - 21 = 26
Then we need to solve the other side 17, 43.
43 - 17 = 26
So this means the answer for A is 26.
To move this along i'm going to solve the others for you. (Starting with B then C)
17 + 36 = 53
100 - 47 = 53
B is 53.
Next C
21 + 36 = 57
100 - 43 = 57
C is 57.
Answer:
an amount of space between two things or people.
Step-by-step explanation:
(a) Let
denote the amount of sugar in the tank at time
. The tank starts with only pure water, so
.
(b) Sugar flows in at a rate of
(0.07 kg/L) * (7 L/min) = 0.49 kg/min = 49/100 kg/min
and flows out at a rate of
(<em>A(t)</em>/1080 kg/L) * (7 L/min) = 7<em>A(t)</em>/1080 kg/min
so that the net rate of change of
is governed by the ODE,
![\dfrac{\mathrm dA(t)}[\mathrm dt}=\dfrac{49}{100}-\dfrac{7A(t)}{1080}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20dA%28t%29%7D%5B%5Cmathrm%20dt%7D%3D%5Cdfrac%7B49%7D%7B100%7D-%5Cdfrac%7B7A%28t%29%7D%7B1080%7D)
or
![A'(t)+\dfrac7{1080}A(t)=\dfrac{49}{100}](https://tex.z-dn.net/?f=A%27%28t%29%2B%5Cdfrac7%7B1080%7DA%28t%29%3D%5Cdfrac%7B49%7D%7B100%7D)
Multiply both sides by the integrating factor
to condense the left side into the derivative of a product:
![e^{\frac{7t}{1080}}A'(t)+\dfrac7{1080}e^{\frac{7t}{1080}}A(t)=\dfrac{49}{100}e^{\frac{7t}{1080}}](https://tex.z-dn.net/?f=e%5E%7B%5Cfrac%7B7t%7D%7B1080%7D%7DA%27%28t%29%2B%5Cdfrac7%7B1080%7De%5E%7B%5Cfrac%7B7t%7D%7B1080%7D%7DA%28t%29%3D%5Cdfrac%7B49%7D%7B100%7De%5E%7B%5Cfrac%7B7t%7D%7B1080%7D%7D)
![\left(e^{\frac{7t}{1080}}A(t)\right)'=\dfrac{49}{100}e^{\frac{7t}{1080}}](https://tex.z-dn.net/?f=%5Cleft%28e%5E%7B%5Cfrac%7B7t%7D%7B1080%7D%7DA%28t%29%5Cright%29%27%3D%5Cdfrac%7B49%7D%7B100%7De%5E%7B%5Cfrac%7B7t%7D%7B1080%7D%7D)
Integrate both sides:
![e^{\frac{7t}{1080}}A(t)=\displaystyle\frac{49}{100}\int e^{\frac{7t}{1080}}\,\mathrm dt](https://tex.z-dn.net/?f=e%5E%7B%5Cfrac%7B7t%7D%7B1080%7D%7DA%28t%29%3D%5Cdisplaystyle%5Cfrac%7B49%7D%7B100%7D%5Cint%20e%5E%7B%5Cfrac%7B7t%7D%7B1080%7D%7D%5C%2C%5Cmathrm%20dt)
![e^{\frac{7t}{1080}}A(t)=\dfrac{378}5e^{\frac{7t}{1080}}+C](https://tex.z-dn.net/?f=e%5E%7B%5Cfrac%7B7t%7D%7B1080%7D%7DA%28t%29%3D%5Cdfrac%7B378%7D5e%5E%7B%5Cfrac%7B7t%7D%7B1080%7D%7D%2BC)
Solve for
:
![A(t)=\dfrac{378}5+Ce^{-\frac{7t}{1080}}](https://tex.z-dn.net/?f=A%28t%29%3D%5Cdfrac%7B378%7D5%2BCe%5E%7B-%5Cfrac%7B7t%7D%7B1080%7D%7D)
Given that
, we find
![0=\dfrac{378}5+C\implies C=-\dfrac{378}5](https://tex.z-dn.net/?f=0%3D%5Cdfrac%7B378%7D5%2BC%5Cimplies%20C%3D-%5Cdfrac%7B378%7D5)
so that the amount of sugar at any time
is
![\boxed{A(t)=\dfrac{378}5\left(1-e^{-\frac{7t}{1080}}\right)}](https://tex.z-dn.net/?f=%5Cboxed%7BA%28t%29%3D%5Cdfrac%7B378%7D5%5Cleft%281-e%5E%7B-%5Cfrac%7B7t%7D%7B1080%7D%7D%5Cright%29%7D)
(c) As
, the exponential term converges to 0 and we're left with
![\displaystyle\lim_{t\to\infty}A(t)=\frac{378}5](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bt%5Cto%5Cinfty%7DA%28t%29%3D%5Cfrac%7B378%7D5)
or 75.6 kg of sugar.