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tekilochka [14]
3 years ago
12

Which type of function is shown in the table below

Mathematics
2 answers:
deff fn [24]3 years ago
7 0
Hi! I would say this is exponential. Please lmk if I’m right :)
FinnZ [79.3K]3 years ago
4 0
Not a function ;)) hope this helps
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I need help on this problem please help thanks.
Studentka2010 [4]

Answer:

That would be Point R.

Step-by-step explanation:

The absolute value of -1/2 is 1/2

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21 you stoopid

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Combine into a single logarithm.<br><br> 3log(x+y)+2log(x-y)-log(x^2 +y^2)
seropon [69]

Answer:

3\log _{10}\left(x+y\right)+2\log _{10}\left(x-y\right)-\log _{10}\left(x^2+y^2\right)=\log _{10}\left(\frac{\left(x+y\right)^3\left(x-y\right)^2}{x^2+y^2}\right)

Step-by-step explanation:

Given the expression

3log\left(x+y\right)+2log\left(x-y\right)-log\left(x^2\:+y^2\right)

solving to write into a single logarithm

3log\left(x+y\right)+2log\left(x-y\right)-log\left(x^2\:+y^2\right)

  • \mathrm{Apply\:log\:rule}:\quad \:a\log _c\left(b\right)=\log _c\left(b^a\right)

3\log _{10}\left(x+y\right)=\log _{10}\left(\left(x+y\right)^3\right)

so

=\log _{10}\left(\left(x+y\right)^3\right)+2\log _{10}\left(x-y\right)-\log _{10}\left(x^2+y^2\right)

  • \mathrm{Apply\:log\:rule}:\quad \:a\log _c\left(b\right)=\log _c\left(b^a\right)

2\log _{10}\left(x-y\right)=\log _{10}\left(\left(x-y\right)^2\right)

so

=\log _{10}\left(\left(x+y\right)^3\right)+\log _{10}\left(\left(x-y\right)^2\right)-\log _{10}\left(x^2+y^2\right)

  • \mathrm{Apply\:log\:rule}:\quad \log _c\left(a\right)+\log _c\left(b\right)=\log _c\left(ab\right)

\log _{10}\left(\left(x+y\right)^3\right)+\log _{10}\left(\left(x-y\right)^2\right)=\log _{10}\left(\left(x+y\right)^3\left(x-y\right)^2\right)

so

=\log _{10}\left(\left(x+y\right)^3\left(x-y\right)^2\right)-\log _{10}\left(x^2+y^2\right)

  • \mathrm{Apply\:log\:rule}:\quad \log _c\left(a\right)-\log _c\left(b\right)=\log _c\left(\frac{a}{b}\right)

\log _{10}\left(\left(x+y\right)^3\left(x-y\right)^2\right)-\log _{10}\left(x^2+y^2\right)=\log _{10}\left(\frac{\left(x+y\right)^3\left(x-y\right)^2}{x^2+y^2}\right)

=\log _{10}\left(\frac{\left(x+y\right)^3\left(x-y\right)^2}{x^2+y^2}\right)

Thus,

3\log _{10}\left(x+y\right)+2\log _{10}\left(x-y\right)-\log _{10}\left(x^2+y^2\right)=\log _{10}\left(\frac{\left(x+y\right)^3\left(x-y\right)^2}{x^2+y^2}\right)

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Answer:

x² = 60

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Step-by-step explanation:

•°•°•°•°•°

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Fina spent $83.65 in all.
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