Answer:
f(-2) = -6
Step-by-step explanation:
In the expression, to get f(-2);
we simply substitute the value of x with -2
We have this as;
f(-2) = -4(-2)^2 + 10
f(-2) = -4(4) + 10
f(-2) = -16 + 10
f(-2) = -6
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.
f(x) *g(x) = (x-3)(x+11) = x^2 +11x -3x -33 = x^2 +8x -33
Sorry if its wrong :(
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)