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Vlada [557]
3 years ago
9

Help me out here please! Thanks.

Mathematics
2 answers:
Setler79 [48]3 years ago
4 0

Answer:

A. \dfrac{1}{5} \log_3 x + \log_3 y

Step-by-step explanation:

\log_3(\sqrt[5]{x} \cdot y) =

The log of a product is the sum of the logs.

= \log_3 \sqrt[5]{x} + \log_3 y

Now, write the root as a rational power.

= \log_3 x^\frac{1}{5} + \log_3 y

The log of a power is the the exponent times the log of the base.

= \dfrac{1}{5} \log_3 x + \log_3 y

leva [86]3 years ago
3 0

Answer:

A. \frac{1}{5} log_{3}x +log_{3}y

Step-by-step explanation:

We have the expression

log_{3} (\sqrt[5]{x} *y)

As these two values are being multiplied, we can separate the two and the sum of them will be equal to the multiplied version

log_{3}\sqrt[5]{x} +log_{3}y

The \sqrt[5]{x} can be rewritten as x^{\frac{1}{5} }. This allows us to use the exponent rule. This means that it can be written as

\frac{1}{5} log_{3}x +log_{3}y

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I need help with number 9
ch4aika [34]

Answer:

D

Step-by-step explanation:

We can safely assume we are dealing with an arithmetic scale because the type of scale isn't mentioned anywhere. The more right a point is on the scale, the higher its value is. We are given two points of the scale: 3 and 4. Because the scale is arithmetic, we know that 3.8 must lie on 4/5s of the length between 3 and 4 to the right of 3, which is exactly where point D is.

3 0
3 years ago
Write an inequality to represent each sentence. Use n for the variable.
Zepler [3.9K]

Answer:

17>n is the correct answer

7 0
2 years ago
Read 2 more answers
<img src="https://tex.z-dn.net/?f=%5Cfrac%7B3%2F7%7D%7B7%5Csqrt%7B10%7D%2F2%20%7D" id="TexFormula1" title="\frac{3/7}{7\sqrt{10}
VashaNatasha [74]

Multiply the numerator and denominator by 7×2 = 14 to eliminate the denominators of those fractions:

\dfrac{\dfrac37}{\dfrac{7\sqrt{10}}2}\times\dfrac{14}{14}=\dfrac{3\times2}{7\sqrt{10}\times7}=\dfrac6{49\sqrt{10}}

Rationalize the denominator by multiplying both numerator and denominator by √10:

\dfrac6{49\sqrt{10}}\times\dfrac{\sqrt{10}}{\sqrt{10}}=\dfrac{6\sqrt{10}}{49(\sqrt{10})^2}=\dfrac{6\sqrt{10}}{49\times10}=\dfrac{6\sqrt{10}}{490}

Lastly, cancel the common factor of 2 in both the numerator and denominator (which comes from 6 = 2×3 and 490 = 2×245):

\dfrac{6\sqrt{10}}{490}=\dfrac{3\sqrt{10}}{245}}

8 0
3 years ago
What is mine times nine
statuscvo [17]
If you are meaning Nine times Nine the answer is 81.
8 0
3 years ago
Translate and simply: subtract 18 from -11
navik [9.2K]
Answer: -29

Subtract 18 FROM -11.
That means that 18 would go after -11.

-11 - 18 = -29
4 0
3 years ago
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