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BartSMP [9]
3 years ago
14

What is the solution to the equation below? Round your answer to two decimal places. log5 x=2.1

Mathematics
1 answer:
Natali5045456 [20]3 years ago
5 0

Answer:

D. x=29.37

Step-by-step explanation:

To answer this question you have to understand how log work. Log is the abbreviation of logarithm and it works as an inverse function of exponentiation.  

The rules should be like this:

 

log_{a} b = c\\a^{c} = b\\

Variable a in the equation is called base. If it left empty you can assume its base 10. The log in this question using 5 as a base. Variable b will be x and variable c will be 2.1

Now, let's change the variable of the equation using the information from the question.  

log_{a} b = c\\\log_{5} x = 2.1\\5^{2.1} = x\\x= log_5}^{-1}2.1\\x=29.365473577\\x=29.37

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In august the simons water bill was $48.in September it was 15% lower . What was the simons water bill in September
slega [8]
Water bill of Simons in the month of August = $48
Percentage of bill that was lower in the month of September = 15%
then
Amount of bill that was lower in the month of September = (15/100) * 48
                                                                                           = 720/100 dollars
                                                                                           = 7.2 dollars
So
The bill for the month of September = (48 - 7.2) dollars
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So Simon will have to pay $40.8 as the water bill for the month of September. I hope the procedure is clear enough for you to understand.
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4 years ago
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A line passes through the points (1, 4) and (3, –4). which is the slope of the line?
mr_godi [17]

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4 0
3 years ago
-4x = 16<br><br> a -64<br><br><br><br><br> b 4<br><br><br><br><br> c 20<br><br><br><br><br> d-4
cricket20 [7]

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How many 10-digit ternary strings are there that contain exactly two 0s, three 1s, and five 2s?
Svet_ta [14]

There are \dbinom{10}2 ways of picking 2 of the 10 available positions for a 0. 8 positions remain.

There are \dbinom83 ways of picking 3 of the 8 available positions for a 1. 5 positions remain, but we're filling all of them with 2s, and there's \dbinom55=1 way of doing that.

So we have

\dbinom{10}2\dbinom83\dbinom55=\dfrac{10!}{2!(10-2)!}\dfrac{8!}{3!(8-3)!}\dfrac{5!}{5!(5-5)!}=2520

The last expression has a more compact form in terms of the so-called multinomial coefficient,

\dbinom{10}{2,3,5}=\dfrac{10!}{2!3!5!}=2520

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3 years ago
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saw5 [17]

Answer:

-21, -16, 4, 9

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