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Andre45 [30]
3 years ago
7

Use the power property to rewrite log3x9.

Mathematics
2 answers:
Afina-wow [57]3 years ago
8 0

Answer:

the answer is b on edg. i just did it.

Step-by-step explanation:

posledela3 years ago
4 0
For properties of logarithm we have the following:
 loga (x ^ b) = b * loga (x)
 Therefore, following this property we have for this case:
 log3 (x ^ 9) = 9log3 (x)
 Answer:
 
the power property to rewrite log3x9 is:
 
B) 9log3x
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When a point lies on the x-axis, y = 0, the coordinates are ( x , 0 ).
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Explain why 4 x 2/5 is the same operation as 2/5 + 2/5 +2/5 + 2/5
wlad13 [49]

Answer:

Step-by-step explanation: Multiply both numerators and denominators. Result fraction keep to lowest possible denominator GCD(8, 5)=1. In the next intermediate step the fraction result cannot be further simplified by cancelling. In words - four multiplied by two fifths=eight fifths.

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3 years ago
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3 years ago
Initially a tank contains 10 liters of pure water. Brine of unknown (but constant) concentration of salt is flowing in at 1 lite
zhenek [66]

Answer:

Therefore the concentration of salt in the incoming brine is 1.73 g/L.

Step-by-step explanation:

Here the amount of incoming and outgoing of water are equal. Then the amount of water in the tank remain same = 10 liters.

Let the concentration of salt  be a gram/L

Let the amount salt in the tank at any time t be Q(t).

\frac{dQ}{dt} =\textrm {incoming rate - outgoing rate}

Incoming rate = (a g/L)×(1 L/min)

                       =a g/min

The concentration of salt in the tank at any time t is = \frac{Q(t)}{10}  g/L

Outgoing rate = (\frac{Q(t)}{10} g/L)(1 L/ min) \frac{Q(t)}{10} g/min

\frac{dQ}{dt} = a- \frac{Q(t)}{10}

\Rightarrow \frac{dQ}{10a-Q(t)} =\frac{1}{10} dt

Integrating both sides

\Rightarrow \int \frac{dQ}{10a-Q(t)} =\int\frac{1}{10} dt

\Rightarrow -log|10a-Q(t)|=\frac{1}{10} t +c        [ where c arbitrary constant]

Initial condition when t= 20 , Q(t)= 15 gram

\Rightarrow -log|10a-15|=\frac{1}{10}\times 20 +c

\Rightarrow -log|10a-15|-2=c

Therefore ,

-log|10a-Q(t)|=\frac{1}{10} t -log|10a-15|-2 .......(1)

In the starting time t=0 and Q(t)=0

Putting t=0 and Q(t)=0  in equation (1) we get

- log|10a|= -log|10a-15| -2

\Rightarrow- log|10a|+log|10a-15|= -2

\Rightarrow log|\frac{10a-15}{10a}|= -2

\Rightarrow |\frac{10a-15}{10a}|=e ^{-2}

\Rightarrow 1-\frac{15}{10a} =e^{-2}

\Rightarrow \frac{15}{10a} =1-e^{-2}

\Rightarrow \frac{3}{2a} =1-e^{-2}

\Rightarrow2a= \frac{3}{1-e^{-2}}

\Rightarrow a = 1.73

Therefore the concentration of salt in the incoming brine is 1.73 g/L

8 0
3 years ago
What is the total number of digits required in numbering a book that has 1724 pages?
Reptile [31]

Answer:

5,789

Step-by-step explanation:

I got 5,789 because there are 9 one digit numbers. 1-9 -> 9x1=9. There are 90 two digit numbers. (10-99)-> 90x2=180. There are 900 three digit numbers. (100-999)-> 900x3=2700. There are 725 four digit numbers. (1000-1724)-> 724x4= 2900. Then, I added 9+180+2700+2900=5789.

In conclusion, the total number of digits required in numbering a book that has 1724 pages is 5789.

8 0
3 years ago
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