3 years ago

Is andies solution correct

1 answer:

3 0

The solution is correct.

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klasskru [66]

-25/6. hope this helps!

6 0

3 years ago

denis-greek [22]

Bits? what are bits?

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3 years ago

Vlada [557]

Answer:

Yes

Step-by-step explanation:

3 0

3 years ago

Assoli18 [71]

Answer:

167.27 mg.

Step-by-step explanation:

We have been given that the half-life of Radium-226 is 1590 years and a sample contains 400 mg.

We will use half life formula to solve our given problem.

$N(t)=N_0*(\frac{1}{2})^{\frac{t}{t/2}$ , where N(t)= Final amount after t years, $N_0$ = Original amount, t/2= half life in years.

Now let us substitute our given values in half-life formula.

$N(2000)=400*(\frac{1}{2})^{\frac{2000}{1590}$

$N(2000)=400*(0.5)^{1.2578616352201258}$

$N(2000)=400*0.4181633028874878239$

$N(2000)=167.26532115499512956\approx 167.27$

Therefore, the remaining amount of Radium-226 after 2000 years will be 167.27 mg.

3 0

3 years ago

inn [45]

Since a^2 - b^2 = 1, the value of y = 1 because x=1 and 1^2 - 1^2 = 0

5 0

3 years ago