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Ray Of Light [21]
3 years ago
9

I have no idea how to know when to factor or know when the polynomial is completely factored or not. PLEASE HELP!

Mathematics
1 answer:
Roman55 [17]3 years ago
8 0

Problem 1 is fully factored as each term is a binomial raised to some exponent. If the exponent isn't showing up, it's because it is 1. Recall that x^1 = x.

Problem 2 can be factored further because x^2-8x+16 factors to (x-4)(x-4) or (x-4)^2. To get this factorization, you find two numbers that multiply to 16 and add to -8. Those two numbers are -4 and -4 which is where the (x-4)(x-4) comes from. Overall, the entire thing factors to (x-4)^2*(x+3)*(x-2)

Problem 3 is a similar story. We can factor x^2-1 into (x-1)(x+1). I used the difference of squares rule here. Or you can think of x^2-1 as x^2+0x-1, then find two numbers that multiply to -1 and add to 0. Those two numbers are +1 and -1 which leads to (x+1)(x-1). So the full factorization is (x-1)(x+1)(x+1)(x-4) which is the same as (x-1)(x+1)^2(x-4)

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Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains liters of a dye solution with a
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t = 460.52 min

Step-by-step explanation:

Here is the complete question

Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 liters of a dye solution with a concentration of 1 g/liter. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 liters/min, the well-stirred solution flowing out at the same rate.Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.

Solution

Let Q(t) represent the amount of dye at any time t. Q' represent the net rate of change of amount of dye in the tank. Q' = inflow - outflow.

inflow = 0 (since the incoming water contains no dye)

outflow = concentration × rate of water inflow

Concentration = Quantity/volume = Q/200

outflow = concentration × rate of water inflow = Q/200 g/liter × 2 liters/min = Q/100 g/min.

So, Q' = inflow - outflow = 0 - Q/100

Q' = -Q/100 This is our differential equation. We solve it as follows

Q'/Q = -1/100

∫Q'/Q = ∫-1/100

㏑Q  = -t/100 + c

Q(t) = e^{(-t/100 + c)} = e^{(-t/100)}e^{c}  = Ae^{(-t/100)}\\Q(t) = Ae^{(-t/100)}

when t = 0, Q = 200 L × 1 g/L = 200 g

Q(0) = 200 = Ae^{(-0/100)} = Ae^{(0)} = A\\A = 200.\\So, Q(t) = 200e^{(-t/100)}

We are to find t when Q = 1% of its original value. 1% of 200 g = 0.01 × 200 = 2

2 = 200e^{(-t/100)}\\\frac{2}{200} =  e^{(-t/100)}

㏑0.01 = -t/100

t = -100㏑0.01

t = 460.52 min

6 0
3 years ago
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