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tia_tia [17]
3 years ago
13

Plz help me sovle this4(-2 1/3) + 7 1/6

Mathematics
2 answers:
Radda [10]3 years ago
8 0
4(-21/3)+71/6
-91/3+71/6
-21/6
docker41 [41]3 years ago
4 0

Answer:

-2 1/6

Step-by-step explanation:

4(-2 1/3) + 7 1/6

-9 1/3 + 7 1/6

-2 1/6

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Suppose that $p$ and $q$ are positive numbers for which \[\log_9 p = \log_{12} q = \log_{16} (p + q).\] what is the value of $q/
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Given p,q>0 and \log_9p=\log_{12}q=\log_{16}(p+q)=x (say).

Then,

p=9^x\\ q=12^x\\ p+q=16^x

From the above 3 equations,

\frac{q}{p} =(\frac{12}{9} )^x\\ \frac{q}{p} =(\frac{4}{3} )^x\\ \frac{p+q}{p} =(\frac{16}{9} )^x\\ \frac{p+q}{p} =(\frac{4}{3} )^{2x}\\

From the equations, we get

\frac{p+q}{p}=(\frac{q}{p})^2\\ 1+\frac{q}{p}=(\frac{q}{p})^2\\ (\frac{q}{p})^2-\frac{q}{p}-1=0\\ \frac{q}{p}=\frac{1 \pm \sqrt{5}}{2}

Since p,q>0, the negative value is rejected.

\frac{q}{p}=\frac{1 + \sqrt{5}}{2}

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3 years ago
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