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Rina8888 [55]
3 years ago
12

Increase 48m by 20% ??????????????????????????////

Mathematics
2 answers:
lakkis [162]3 years ago
7 0
You need to increase 48 by 20%.

First, change 20% to a decimal: 0.20

Next, multiply 48*0.20=9.6

Now, add 9.6 to 48.

Your final answer is: 57.6 meters.
Anastaziya [24]3 years ago
6 0
Someone else on Brainly had the same question, he showed the work so I copied it so you can see it.

We have 48 meters.
We need to increase it by 20%.
That means we have 48 and then we increase it by 20%.

That can be expressed as:
48 + (48 * 20%)

Now, another way to write that would be:

48 * 120%

Now, 120% is nothing but 120/100. This can even be simplified to 1.2.

48 * 120%
= 48 * 1.2
= 57.6

48m increased by 20% is 57.6m<span>.</span>
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\Huge \boxed{x=24}

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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
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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