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malfutka [58]
3 years ago
10

During a thunderstorm, 600 millimeters of rain fell in 30 minutes. How fast did the rain fall, in millimeters per minute?

Mathematics
2 answers:
madreJ [45]3 years ago
5 0
20 milliliters per minute
N76 [4]3 years ago
4 0
The rain fell 20 millimeters per minute.

Explanation:

To find the answer, we must divide the amount of rain by the number of minutes it took to fall.

600 divided by 30 = 20 millimeters.
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Multiply the first number by its reciprocal
Tanzania [10]

Answer:

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6 0
2 years ago
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Express this to single logarithm
zzz [600]

Answer:  \log_{2}\left(\frac{q^2\sqrt{m}}{n^3}\right)

We have something in the form log(x/y) where x = q^2*sqrt(m) and y = n^3. The log is base 2.

===========================================================

Explanation:

It seems strange how the first two logs you wrote are base 2, but the third one is not. I'll assume that you meant to say it's also base 2. Because base 2 is fundamental to computing, logs of this nature are often referred to as binary logarithms.

I'm going to use these three log rules, which apply to any base.

  1. log(A) + log(B) = log(A*B)
  2. log(A) - log(B) = log(A/B)
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From there, we can then say the following:

\frac{1}{2}\log_{2}\left(m\right)-3\log_{2}\left(n\right)+2\log_{2}\left(q\right)\\\\\log_{2}\left(m^{1/2}\right)-\log_{2}\left(n^3\right)+\log_{2}\left(q^2\right) \ \text{ .... use log rule 3}\\\\\log_{2}\left(\sqrt{m}\right)+\log_{2}\left(q^2\right)-\log_{2}\left(n^3\right)\\\\\log_{2}\left(\sqrt{m}*q^2\right)-\log_{2}\left(n^3\right) \ \text{ .... use log rule 1}\\\\\log_{2}\left(\frac{q^2\sqrt{m}}{n^3}\right) \ \text{ .... use log rule 2}

8 0
2 years ago
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IrinaK [193]
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EXPLANATION:
9+x/6=11
(54+x)/6=11 (finding common denominator)
x/6=11-54/6 (rearranging terms)
x/6=(66-54)/6 (finding common denominator so 6 gets cancelled from both the sides)
x=12
6 0
2 years ago
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How many centimeters are in 3.2 inches? [1 inch = 2.5 cm]
frez [133]
2.54 cm = 1 inch

3.2 inches * 2.54 cm / inch =8.128 centimeters



7 0
3 years ago
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