3 years ago

Christie's Auction sold a painting for

Mathematics

1 answer:

Tems11 [23]3 years ago

5 0

24,500$. 100%
X$. 15%
X=24,000•15/100
X=3,675 premium

24,000-3,675=20,825$ bidder pay Christine.

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Question 1 ( point) A 64g sample of Germanium-66 is left undisturbed for 12.5 hours. At the end of that period, only 2.0g remain

iVinArrow [24]

Answer:

2.5 half-lives

Step-by-step explanation:

Find $t_{h}$ in $N(t)=N_{0}e^{-kt}$ given $N(t) = 2, N_{0} =64,t=\frac{25}{2}$

$N(t)$ is the amount after the time $t$ , $N_0$ `is the initial amount, $t_{h}$ is the half-life

We know that after half-life there will be twice less the initial quantity:

$N(t_{h})=\frac{N_{0} }{2}=N_{0}e^{kt_{h} }$ .

Simplifying gives $\frac{1}{2} =e^{-kt_h} or k = -\frac{In(\frac{1}{2}) }{t_{h} } .$

Plugging this into the initial equation, we obtain that $N(t)=N_{0}e^\frac{In(\frac{1}{2}) }{t_{h} }$ or $N(t)= N_{0} (\frac{1}{2})^\frac{t}{t_h}.$

Finally, just plug in the given values and find the unknown one.

From $2=64(\frac{1}{2})^\frac{\frac{25}{2} }{t_h}$ , we have that $t_h$ = $\frac{25 In (2)}{2 In (32)\\}$

So you should get: $t_{h} =\frac{25 In (2)}{2 In (32)\\}=2.5.$

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The answer of 8+11=40

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Step-by-step explanation:

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Answer:

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30-2(7+2)-1

PEMDAS says parentheses first

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nevsk [136]

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