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MrRa [10]
3 years ago
10

Find the distance between 6.2 and -4.3 on a number line.

Mathematics
1 answer:
Nitella [24]3 years ago
8 0

Answer:

The two given numbers are 10.5 units apart.

Step-by-step explanation:

Subtract the smaller from the larger.  The larger number here is 6.2. Then we have

6.2 - (-4.3) = 6.2 + 4.3 = 10.5

The two given numbers are 10.5 units apart.

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In a choir, the ratio of boys to girls is 5 to 6. There are 10 boys in the choir. What is the total number of boys and girls in
Aleks [24]

Answer:

5:6 --> 10:12

Step-by-step explanation:

A good way to think of this is to provide a perspective: there are 5 boys every time there is 6 girls. So if there is 10 boys, that means there would be 12 girls (because 5 *2 = 10 and 6*2 = 12).

tl:dr: 5 *2 = 10 so 6*2 = 12

7 0
2 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
Can u please help me i dont get it please thank u soo much
Bumek [7]

12pm: -3° + 4° = 1°

7pm: 1° - 5° = -4°

the answer is C.

7 0
3 years ago
Please someone explain this one? Thank you!
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About 7,460 punds if you add everything together

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