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Colt1911 [192]
3 years ago
9

What do 15/43 = in simplest form

Mathematics
2 answers:
alina1380 [7]3 years ago
5 0
Idk I’m sorryyy I’m having trouble tooo
antoniya [11.8K]3 years ago
3 0

Answer:

0.348

Step-by-step explanation:

The fraction 15/43 is already in the simplest form, so it isn't possible to reduce it any further.

This is a PROPER FRACTION once the absolute value of the top number or numerator (15) is smaller than the absolute value of the bottom number or denomintor (43).

The fraction 15/43 is equal to 15÷43 and can also be expressed in decimal form as 0.348837.

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Step-by-step explanation:

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7 0
2 years ago
Simplify the expression below and write it as a single logarithm:
OLEGan [10]

The simplification of 3log(x + 4) – 2log(x – 7) + 5log(x - 2) - log(x^2) is \log \left(\frac{(x+4)^{3} \times(x-2)^{5}}{(x-7)^{2} \times x^{2}}\right)

<u>Solution:</u>

Given, expression is 3 \log (x+4)-2 \log (x-7)+5 \log (x-2)-\log \left(x^{2}\right)

We have to write in as single logarithm by simplifying it.

Now, take the given expression.

\rightarrow 3 \log (x+4)-2 \log (x-7)+5 \log (x-2)-\log \left(x^{2}\right)

Rearranging the terms we get,

\left.\rightarrow 3 \log (x+4)+5 \log (x-2)-2 \log (x-7)+\log \left(x^{2}\right)\right)

\text { since a } \times \log b=\log \left(b^{a}\right)

\rightarrow \log (x+4)^{3}+\log (x-2)^{5}-\left(\log (x-7)^{2}+\log \left(x^{2}\right)\right)

\text { We know that } \log a \times \log b=\log a b

\rightarrow \log \left((x+4)^{3} \times(x-2)^{5}\right)-\left(\log \left((x-7)^{2} \times\left(x^{2}\right)\right)\right.

\text { We know that } \log a-\log b=\log \frac{a}{b}

\rightarrow \log \left(\frac{(x+4)^{3} \times(x-2)^{5}}{(x-7)^{2} \times x^{2}}\right)

Hence, the simplified form \rightarrow \log \left(\frac{(x+4)^{3} \times(x-2)^{5}}{(x-7)^{2} \times x^{2}}\right)

4 0
3 years ago
Read 2 more answers
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