Answer:
The equation of line is y = 3 x - 1 , i.e option A .
The rate of change = m = 3
Step-by-step explanation:
Given as :
The points are
(
,
) = (- 2, -7)
(
,
) = (1 , 2)
Let The slope = m
<u>Now, The slope is calculated in points form</u>
So, Slope = 
I.e m = 
or, m = 
I.e m = 3
So, The slope of points = m = 3
I,e Rate of change = m = 3
<u>Now, the equation of line in slope - point form can be written as</u>
y - = m ( x - )
where m is the slope of line
i.e y - (-7) = ( 3 ) × ( x - (-2) )
or, y + 7 = 3 × (x + 2)
∴ y + 7 = 3 x + 6
Or, y = 3 x + 6 - 7
Or, y = 3 x - 1
So, The equation of line is y = 3 x - 1
Hence, The equation of line is y = 3 x - 1 , i.e option A .
And The rate of change = m = 3 Answer
The answer to your question is 2.5 hope this helps
Answer:
20
Step-by-step explanation: add them all together
At the start, the tank contains
(0.02 g/L) * (1000 L) = 20 g
of chlorine. Let <em>c</em> (<em>t</em> ) denote the amount of chlorine (in grams) in the tank at time <em>t </em>.
Pure water is pumped into the tank, so no chlorine is flowing into it, but is flowing out at a rate of
(<em>c</em> (<em>t</em> )/(1000 + (10 - 25)<em>t</em> ) g/L) * (25 L/s) = 5<em>c</em> (<em>t</em> ) /(200 - 3<em>t</em> ) g/s
In case it's unclear why this is the case:
The amount of liquid in the tank at the start is 1000 L. If water is pumped in at a rate of 10 L/s, then after <em>t</em> s there will be (1000 + 10<em>t</em> ) L of liquid in the tank. But we're also removing 25 L from the tank per second, so there is a net "gain" of 10 - 25 = -15 L of liquid each second. So the volume of liquid in the tank at time <em>t</em> is (1000 - 15<em>t </em>) L. Then the concentration of chlorine per unit volume is <em>c</em> (<em>t</em> ) divided by this volume.
So the amount of chlorine in the tank changes according to
which is a linear equation. Move the non-derivative term to the left, then multiply both sides by the integrating factor 1/(200 - 5<em>t</em> )^(5/3), then integrate both sides to solve for <em>c</em> (<em>t</em> ):
![\dfrac{\mathrm d}{\mathrm dt}\left[\dfrac{c(t)}{(200-3t)^{5/3}}\right]=0](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dt%7D%5Cleft%5B%5Cdfrac%7Bc%28t%29%7D%7B%28200-3t%29%5E%7B5%2F3%7D%7D%5Cright%5D%3D0)
There are 20 g of chlorine at the start, so <em>c</em> (0) = 20. Use this to solve for <em>C</em> :
![\implies\boxed{c(t)=\dfrac1{200}\sqrt[3]{\dfrac{(200-3t)^5}5}}](https://tex.z-dn.net/?f=%5Cimplies%5Cboxed%7Bc%28t%29%3D%5Cdfrac1%7B200%7D%5Csqrt%5B3%5D%7B%5Cdfrac%7B%28200-3t%29%5E5%7D5%7D%7D)
