Answer:
A=163.66
Area = Length times width
so 33.4 times 4.9
Answer:
If the scale factor is represented by , then the area of the scale drawing is
times the corresponding area of the original drawing.
Step-by-step explanation:
If the scale factor is represented by , then the area of the scale drawing is
times the corresponding area of the original drawing.
For example:
<em>Given a scale factor of 25%, would the quotient of the area of the scale drawing to the area of the original drawing be? </em>
<u>Answer:</u>
The quotient of the areas would be equal to the square of the scale factor. Therefore, the quotient of the scale drawing to the original in this example would be equal to
Hope it will find you well.
(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.
You’re answer is 4
3-7(4)=-25 so f=4
3x+2x=80
5x=80
x=8
*there should be a degree sign after but i can’t type that*