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)
Hello!
3/4 = 0.75 as a decimal
We know that Jen picks 0.75 of a gallon of strawberries in half an hour, and now we want to find out how many gallons she'll pick in 2 hours.
Half an hour = 0.5
2 ÷ 0.5 = 4 There are four half hours in 2 hours.
0.75 × 4 = 3
ANSWER:
Jen will have picked 3 gallons of strawberries in 2 hours.
Change of y / over change of x
Go over 4, up 45
B is your correct answer
Hope this helps!
<span>I note that this problem starts out with "Which is a factor of ... " This implies that you were given several answer choices. If that's the case, it's unfortunate that you haven't shared them.
I thought I'd try finding roots of this function using synthetic division. See below:
f(x) = 6x^4 – 21x^3 – 4x^2 + 24x – 35
Please use " ^ " to denote exponentiation. Thanks.
Possible zeros of this poly are factors of 35: plus or minus 1, plus or minus 5, plus or minus 7. Use synthetic division; determine whether or not there is a non-zero remainder in each case. If none of these work, form rational divisors from 35 and 6 and try them: 5/6, 7/6, 1/6, etc.
Provided that you have copied down the function
</span>f(x) = 6x^4 – 21x^3 – 4x^2 + 24x – 35 properly, this approach will eventually turn up 1 or 2 zeros of this poly. Obviously it'd be much easier if you'd check out the possible answers given you with this problem.
By graphing this function, I found that the graph crosses the x-axis at 7/2. There is another root.
Using synth. div. to check whether or not 7/2 is a root:
___________________________
7/2 / 6 -21 -4 24 -35
21 0 -14 35
----------- ------------------------------
6 0 -4 10 0
Because the remainder is zero, 7/2 (or 3.5) is a root of the polynomial. Thus, (x-3.5), or (x-7/2), is a factor.
I hope this helps you
-2a^2b.4.a^5.b^2
-8.(a^2+5)(b^1+2)
-8.a^7.b^3
