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4 years ago

11

If i have ten fish and six drawn and two come with the left of the fish that didn't drawn but one dies how many fish do i have ?

????????????
please get back to me

-thank you

2 answers:

olga_2 [115]4 years ago

8 0

11 i think im wrong thought .

Paraphin [41]4 years ago

8 0

11 i think im not sure hopw it helps lolololoololo

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Considering the expression

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$\ln \left(e^{\ln \left(x\right)}\right)=\ln \left(x\right),\:\space\ln \left(e^{\ln \left(x\right)2}\right)=\ln \left(x\right)2$

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$3\ln \left(x\right)=2\ln \left(8\right)$

$\mathrm{Divide\:both\:sides\:by\:}3$

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$\frac{2\ln \left(8\right)}{3}$

As

$\ln \left(8\right):\quad 3\ln \left(2\right)$

Because

$\ln \left(8\right)$

$\mathrm{Rewrite\:}8\mathrm{\:in\:power-base\:form:}\quad 8=2^3$

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$\mathrm{Apply\:log\:rule}:\quad \log _a\left(x^b\right)=b\cdot \log _a\left(x\right)$

$\ln \left(2^3\right)=3\ln \left(2\right)$

So

$\frac{2\ln \left(8\right)}{3}=\frac{2\cdot \:3\ln \left(2\right)}{3}$

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So, equation A becomes

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Therefore, the solution is

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