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goldfiish [28.3K]
3 years ago
9

HALP PLS NO LINKS OR FILES PLS SCREENSHOT BELOW!!!

Mathematics
1 answer:
Sidana [21]3 years ago
7 0

Answer:

12 and 13

Step-by-step explanation:

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HELP IMMEDIATELY, ILL GIVE BRAINLY!!!!!!
klio [65]

Step-by-step explanation:

nodndnsjsjjsjsjjsjsjsjjsjdjjdjdkd

7 0
2 years ago
Simplify 13 to the second power pretty please :3
just olya [345]

13 to the second power is 169.

Hope this helps!

6 0
2 years ago
What percent is 608 million of 845 million? Round to the nearest 100th place and Enter your answer with the percent symbol.
bezimeni [28]

Answer: 72%

Step-by-step explanation: Well, we want to know what percent 608 is of 845, so to solve that, let's call 845 100%, as we're solving it terms of 845. 845 is 100%, then it follows that 8.45 is 1%.

Now we can see how many 8.45's go into 608, which comes out to 71.95266...%, which can be rounded nicely to 72%

8 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
Please help this is 5th Grade Math also show work 15 Points awarded
Dimas [21]

Answer:

The only one that would be reasonable is the first one...and maybe the third one too

Step-by-step explanation:


8 0
3 years ago
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