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algol [13]
3 years ago
10

Please help it’s over due

Mathematics
2 answers:
AlladinOne [14]3 years ago
7 0

It would be 0.93, 1.02, 1.11

Diano4ka-milaya [45]3 years ago
6 0
Your increasing by .09 each time so the next 3 would be .93, 1.02 and 1.11.
so the rule would be x+.09=y
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Please help logarithms!
nlexa [21]

Given:

\log_34\approx 1.262

\log_37\approx 1.771

To find:

The value of \log_3\left(\dfrac{4}{49}\right).

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\log_34\approx 1.262

\log_37\approx 1.771

Using properties of log, we get

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\log_3\left(\dfrac{4}{49}\right)=\log_34-\log_37^2      

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\log_3\left(\dfrac{4}{49}\right)=1.262-2(1.771)

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