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3 years ago

6

684,658 round to the nearest hundred thousand

2 answers:

OverLord2011 [107]3 years ago

8 0

700,000 is the answer I believe

Gennadij [26K]3 years ago

8 0

700,000 is the answer because you will have to round nearest the hundred thousand so the closest is 700,000 thousand hundred

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Arturiano [62]

Answer:

t = 6162 years

Step-by-step explanation:

Apparently you are asked to use the constant k = 0.0001 instead of the most common 0.00012. We'll answer the problem using such number you typed, but make sure that there is no omission in your typing.

The equation to use is that for carbon decay:

$N(t)=N_0\,e^{-0.0001\,\,t}$

If the skull contains 54% of its original amount of C14, then that means that :

$N(t)=0.54 \,\,N_0$

we use this to solve for the time (t) in our equation:

$0.54\,\,N_0=N_0\,e^{-0.0001\,\,t}\\\frac{0.54\,\,N_0}{N_0}=e^{-0.0001\,\,t}\\0.54=e^{-0.0001\,\,t}\\ln(0.54)=-0.0001\,\,t\\t=\frac{ln(0.54)}{-0.0001} \\t=6161.86\,\,years$

which rounded to the nearest year as requested gives :

t = 6162 years

8 0

3 years ago

Pls answer these questions

Mila [183]

Answer:

I believe the answers are

B) 1

D) $1,000

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

you ordered a laptop from a company in alaska. The laptop costs $150. they charge you 15% off the cost of the laptop to ship it.

monitta

Answer:

$127.5

Step-by-step explanation:

150*15%=22.5.

150-22.5=127.5

Don't buy this, totally not worth it.

8 0

3 years ago

Read 2 more answers

Vlada [557]

The equation of the line f(x) slope 3

5 0

2 years ago

A particle is moving along the x-axis so that its position at t ≥ 0 is given by s(t)=(t)ln(5t). Find the acceleration of the par

lyudmila [28]

Answer:

$a(\frac{1}{5e})=5e$

Step-by-step explanation:

we are given equation for position function as

$s(t)=tln(5t)$

Since, we have to find acceleration

For finding acceleration , we will find second derivative

$s'(t)=\frac{d}{dt}\left(t\ln \left(5t\right)\right)$

$=\frac{d}{dt}\left(t\right)\ln \left(5t\right)+\frac{d}{dt}\left(\ln \left(5t\right)\right)t$

$=1\cdot \ln \left(5t\right)+\frac{1}{t}t$

$s'(t)=\ln \left(5t\right)+1$

now, we can find derivative again

$s''(t)=\frac{d}{dt}\left(\ln \left(5t\right)+1\right)$

$=\frac{d}{dt}\left(\ln \left(5t\right)\right)+\frac{d}{dt}\left(1\right)$

$=\frac{1}{t}+0$

$a(t)=\frac{1}{t}$

Firstly, we will set velocity =0

and then we can solve for t

$v(t)=s'(t)=\ln \left(5t\right)+1=0$

we get

$t=\frac{1}{5e}$

now, we can plug that into acceleration

and we get

$a(\frac{1}{5e})=\frac{1}{\frac{1}{5e}}$

$a(\frac{1}{5e})=5e$

5 0

3 years ago