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Yuri [45]
3 years ago
5

Hey I need this answered i have no clue what to put pls help ​

Mathematics
1 answer:
Molodets [167]3 years ago
4 0

Answer:

b is correct for Maria but c is correct for the boy

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Solve: k /8 + 1 − = −19
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Astronomers measure large distances in light-years. One light-year is the distance that light can travel in one year, or approxi
Semenov [28]

Answer:

Option B - =8.4672\times 10^{13}

Step-by-step explanation:

Given : Astronomers measure large distances in light-years. One light-year is the distance that light can travel in one year, or approximately 5,880,000,000,000 miles.Suppose a star is 14.4 light years from earth.

To find : In scientific notation, how many miles away a star is from earth?

Solution :

One light-year is the distance that light can travel in one year is = 5,880,000,000,000 miles.

In scientific notation, 5.88\times 10^{12} miles.

In 14.4 lights year the distance is  =14.4\times5.88\times 10^{12}

In 14.4 lights year the distance is  =84.672\times 10^{12}

                                                  or  =8.4672\times 10^{13}

Therefore, Option B is correct.

5 0
3 years ago
A tank contains 1000 L of pure water. Brine that contains 0.05 kg of salt per liter of water enters the tank at a rate of 5 L/mi
Margaret [11]

Answer:

a) y(t)=0.65\frac{Kg}{min}(tmin)

b) y(40)=26Kg

Step-by-step explanation:

Data

Brine a (Ba)

V_{Ba}=5\frac{Lt}{min}\\  Concentration(Bca)=0.05\frac{Kg}{Lt}

Brine b (Bb)

V_{Bb}=10\frac{Lt}{min}\\  Concentration(Bcb)=0.04\frac{Kg}{Lt}

we have that per every minute the amount of solution that enters the tank is the same as the one that leaves the tank (15 Lt / min)

, then the amount of salt (y) left in the tank after (t) minutes: y=V_{Ba}*B_{ca}+V_{Bb}*B_{cb}=5\frac{Lt}{min}*0.05\frac{Kg}{Lt}+10\frac{Lt}{min}*0.04\frac{Kg}{Lt}=\\0.25\frac{Kg}{min}+0.4\frac{Kg}{min}=0.65\frac{Kg}{min}

Finally:

a) y(t)=0.65\frac{Kg}{min}(tmin)

b) y(40)=0.65\frac{Kg}{min}(40min)=26Kg

being y(t) the amount of salt (y) per unit of time (t)

5 0
2 years ago
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