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)
Answer:
See proof below
Step-by-step explanation:
We will use properties of inequalities during the proof.
Let
. then we have that
. Hence, it makes sense to define the positive number delta as
(the inequality guarantees that these numbers are positive).
Intuitively, delta is the shortest distance from y to the endpoints of the interval. Now, we claim that
, and if we prove this, we are done. To prove it, let
, then
. First,
then
hence
On the other hand,
then
hence
. Combining the inequalities, we have that
, therefore
as required.
Answer:
6x-15
Step-by-step explanation:
Answer:
Step-by-step explanation:
(1). Maryam's answer is correct.
(2). x = 2
To expand the given expression we proceed as follows:
(6x²-2x-6)(8x²+7x+8)
=6x²(8x²+7x+8)-2x(8x²+7x+8)-6(8x²+7x+8)
=48x⁴+42x³+48x²-16x³-14x²-16x-48x²-42x-48
putting like terms together:
48x⁴+(42x³-16x³)+(48x²-48x²)+(-16x-42x)-48
=48x⁴+26x³+0x²-58x-48
hence the answer is:
48x⁴+26x³-58x-48