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inessss [21]
3 years ago
14

What is the main goal of debating an issue? Why?

Mathematics
1 answer:
lana66690 [7]3 years ago
7 0
1. Why?= Debate is a method of formally creating an argument in a disciplined manner.

2. What is the Goal?= The goal is to decide on one of the many issues that are being discussed at that time.They are often followed through by voting.

People like to debate so that their opinion is herd , that's my opinion. 

did I help you understand your question? 
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HELP PLEASE THIS IS TIMED BROOOOO
slamgirl [31]
75-22= 53 so the 3rd answer
3 0
3 years ago
5)
cricket20 [7]
The answer is no solution because when you solve the equation

3(x+4)=3x+4

3x +12=3x +4
-3x -3x

You get an answer of 12=4 which is not true so there cannot be a solution to this equation

The answer is A no solution
8 0
3 years ago
What is the average velocity of a car that travels 50 km in 2 hrs?
Alisiya [41]

Step-by-step explanation:

the average velocity of a car that travels 50 km in 2 hrs = 50/2 = 25 km/hrs

6 0
3 years ago
For the situation, write and solve an equation. Then, explain what your solution means. If you get stuck, try drawing a diagram
stellarik [79]

Answer:

64 = 4x + 7        

3 0
3 years ago
Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains liters of a dye solution with a
Alja [10]

Answer:

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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