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3 years ago

What is 1.61 nearest tenth

Mathematics

2 answers:

Kamila [148]3 years ago

5 0

Tenths is the 6 So 1 is below 5 so u round down. 1.6 is the answer Hope I helped

pishuonlain [190]3 years ago

5 0

The nearest tenth for 1.61 is 1.60 because if its 5 or more is greater and 4 or less is smaller

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he radioactive element carbon-14 has a half-life of 5750 years. A scientist determined that the bones from a mastodon had lost

elena-s [515]

Answer:

The bones were 12,485 years old at the time they were discovered.

Step-by-step explanation:

Amount of the element:

The amount of the element after t years is given by the following equation, considering the decay rate proportional to the amount present:

$A(t) = A(0)e^{-kt}$

In which A(0) is the initial amount and k is the decay rate, as a decimal.

The radioactive element carbon-14 has a half-life of 5750 years.

This means that $A(5750) = 0.5A(0)$ , and we use this to find k. So

$A(t) = A(0)e^{-kt}$

$0.5A(0) = A(0)e^{-5750k}$

$e^{-5750k} = 0.5$

$\ln{e^{-5750k}} = \ln{0.5}$

$-5750k = \ln{0.5}$

$k = -\frac{\ln{0.5}}{5750}$

$k = 0.00012054733$

$A(t) = A(0)e^{-0.00012054733t}$

A scientist determined that the bones from a mastodon had lost 77.8 % of their carbon-14. How old were the bones at the time they were discovered?

Had 100 - 77.8 = 22.2% remaining, so this is t for which:

$A(t) = 0.222A(0)$

Then

$0.222A(0) = A(0)e^{-0.00012054733t}$

$e^{-0.00012054733t} = 0.222$

$\ln{e^{-0.00012054733t}} = \ln{0.222}$

$-0.00012054733t = \ln{0.222}$

$t = -\frac{\ln{0.222}}{0.00012054733}$

$t = 12485$

The bones were 12,485 years old at the time they were discovered.

8 0

3 years ago

A football is kicked from the ground and follows the graph of the function y = –t2 + 3t, where y represents height and t represe

vesna_86 [32]

Answer:

The height of the objects are the same after 2 seconds.

Step-by-step explanation:

In order to calculate at which time both objects have the same height we need to find the value of t that makes both equations equal. Therefore:

$-t^2 + 3t = -t + 4\\t^2 - 3t - t + 4 = 0\\t^2 - 4t + 4 = 0\\t_{1,2} = \frac{-(-4) \pm \sqrt{(-4)^2 - 4*1*4}}{2*1} = \frac{4 \pm \sqrt{16 - 16}}{2}\\t_{1,2} = \frac{4}{2} = 2$