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2 years ago

Blake is 7 years older than Heather. In 7 years the sum of their ages will be 55. How old is Blake now?

Mathematics

1 answer:

Andreas93 [3]2 years ago

4 0

Answer:

24 years old

Step-by-step explanation:

Let the age of Blake be x and the age of Heather by y then...

x=7+y

7+x+7+y=55

Notice we are adding additional 7s because the question states "in 7 years"

Simplify the second equation...

7+x+7+y=55

14+x+y=55

Subtract 21 from both sides

x+y=41

Solve using substitution (substitute 7+y for x since x = 7+y)

7+y+y=41

Combine like terms

2y+7=41

Subtract 7 from both sides

2y=34

Divide both sides by 2

y=17

x=7+17=24

Blake is 24 years old now

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Remark work with the two numbers that are 10^4 first. Then add in the one that goes to 10^3

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Method two
Keep the scientific notation.
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Either method works.

You could use your calculator to get the right answer. I'm not sure all calculators retain the scientific notation. Test yours to see what it does. Enter 8.3*10y^x 4 *1 =

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3 years ago

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Answer:

Carbon-14 will take 19,035 years to decay to 10 percent.

Step-by-step explanation:

Exponential Decay Function

A radioactive half-life refers to the amount of time it takes for half of the original isotope to decay.

An exponential decay can be described by the following formula:

$N(t)=N_{0}e^{-\lambda t}$

Where:

No = The quantity of the substance that will decay.

N(t) = The quantity that still remains and has not yet decayed after a time t

$\lambda$ = The decay constant.

One important parameter related to radioactive decay is the half-life:

$\displaystyle t_{1/2}=\frac {\ln(2)}{\lambda }$

If we know the value of the half-life, we can calculate the decay constant:

$\displaystyle \lambda=\frac {\ln(2)}{ t_{1/2}}$

Carbon-14 has a half-life of 5,730 years, thus:

$\displaystyle \lambda=\frac {\ln(2)}{ 5,730}$

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The equation of the remaining quantity of Carbon-14 is:

$N(t)=N_{0}e^{-0.00012097\cdot t}$

We need to calculate the time required for the original amout to reach 10%, thus N(t)=0.10No

$0.10N_o=N_{0}e^{-0.00012097\cdot t}$

Simplifying:

$0.10=e^{-0.00012097\cdot t}$

Taking logarithms:

$ln 0.10=-0.00012097\cdot t$

Solving for t:

$\displaystyle t=\frac{log 0.10}{-0.00012097}$

$t\approx 19,035\ years$

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