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Blababa [14]
3 years ago
13

Solve Highschool Pre cal

Mathematics
1 answer:
Bezzdna [24]3 years ago
6 0

Option C:

\ln 64 y^{2}

Solution:

Given expression: 2 ln 8 + 2 ln y

Applying log rule a \log b=\log b^{a}  in the above expression.  

⇒ 2 \ln 8+2 \ln y=\ln 8^{2}+\ln y^{2}

Applying log rule \log a+\log b=\log (a b) in the above expression.

⇒ \ln 8^{2}+\ln y^{2}=\ln \left(8^{2} y^{2}\right)

We know that 8^{2}=64

⇒ \ln \left(8^{2} y^{2}\right)=\ln 64 y^{2}

Hence, the expression in single natural logarithm is \ln 64 y^{2}.

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Step-by-step explanation:

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x = 63

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<img src="https://tex.z-dn.net/?f=12x%20%5E%7B2%7D%20%20%2B%204x%20%2B%2020" id="TexFormula1" title="12x ^{2} + 4x + 20" alt="1
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<h3>Answer: </h3>

The GCF is 4

The polynomial factors to 4(3x^2+x+5)

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

Further explanation:

Ignore the x terms

We're looking for the GCF of 12, 4 and 20

Factor each to their prime factorization. It might help to do a factor tree, but this is optional.

  • 12 = 2*2*3
  • 4 = 2*2
  • 20 = 2*2*5

Each factorization involves "2*2", which means 2*2 = 4 is the GCF here.

We can then factor like so

12x^2 + 4x + 20\\\\4*3x^2 + 4*x + 4*5\\\\4(3x^2 + x + 5)

The distributive property pulls out that common 4. We can verify this by distributing the 4 back in, so we get the original expression back again.

The polynomial inside the parenthesis cannot be factored further. Proof of this can be found by looking at the roots and noticing that they aren't rational numbers (use the quadratic formula).

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