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Natalka [10]
3 years ago
12

How do I solve 5 + 6y = 9y + 2 + 3(1 - y)

Mathematics
1 answer:
Feliz [49]3 years ago
4 0
You use order of operations

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Teresa drove 936 miles in 13 hours. At the same rate, how many miles would she drive in 9 hours?
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If Teresa drove 936 miles in 13 hours, this would mean she would have driven 72 miles every hour. 72 times 9 equals 648. After 9 hours, she would have driven 648 miles. 
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3 years ago
20 What decimal does this model represent? Explain.
umka2103 [35]

Answer:

The model represents the decimal 73.

Step-by-step explanation:

The model is 10x10. You must count the blue blocks. The number of blue blocks is 73, therefore it represents the decimal 73. (Hope this helped!)

8 0
3 years ago
Find the area of this shape below
rjkz [21]

Answer: 28                      

Step-by-step explanation:

2 × 8 = 16

8 + 4 = 12

12 ÷ 2 = 6

6 × 2 = 12

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6 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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