So who's going to do this aka the queen / King Master
1 answer:
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Answer:
200 mg sodium is in 1 oz of chips and 50 mg sodium is in 1 cup of soda.
Step-by-step explanation:
Let x mg sodium is in 1 oz of chips and and y mg is in 1 cup of soda.
∵ Bryan ate 3 oz of chips and drank 2 cups of soda for a total of 700 mg of sodium.
i.e. 3x + 2y = 700 --------(1),
Jadyn ate 1 oz of chips and drank 3 cups of soda for a total of 350 mg of sodium.
i.e. x + 3y = 350 ---------(2),
Equation (1) - 3 × equation (2),
We get,
2y - 9y = 700 - 1050
-7y = -350
From equation (1),
3x + 2(50) = 700
3x + 100 = 700
3x = 700 - 100
3x = 600
Hence, 200 mg sodium is in 1 oz of chips and 50 mg sodium is in 1 cup of soda.
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> ):
There are 20 g of chlorine at the start, so <em>c</em> (0) = 20. Use this to solve for <em>C</em> :