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2 years ago

6464 is 444 times the difference between Sarah's age, aaa, and 444444. Assume Sarah is older than 44

Mathematics

1 answer:

xxMikexx [17]2 years ago

5 0

Answer:

60 years

Step-by-step explanation:

Given that :

Sarah's age = a

Assume Sarah is older than 44 ; then a > 44

64 = 4 times difference between a and 44

Expressing the problem mathematically ;

64 = 4 * (a - 44)

64 = 4a - 176

64 + 176 = 4a

240 = 4a

a = 240 / 4

a = 60

Hence, Sarah is aged 60

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Step-by-step explanation:

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2 years ago

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2 years ago

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2 years ago

Read 2 more answers

Use the exponential decay model, Upper A equals Upper A 0 e Superscript kt, to solve the following. The half-life of a certai

Akimi4 [234]

Answer:

It will take 7 years ( approx )

Step-by-step explanation:

Given equation that shows the amount of the substance after t years,

$A=A_0 e^{kt}$

Where,

$A_0$ = Initial amount of the substance,

If the half life of the substance is 19 years,

Then if t = 19, amount of the substance = $\frac{A_0}{2}$ ,

i.e.

$\frac{A_0}{2}=A_0 e^{19k}$

$\frac{1}{2} = e^{19k}$

$0.5 = e^{19k}$

Taking ln both sides,

$\ln(0.5) = \ln(e^{19k})$

$\ln(0.5) = 19k$

$\implies k = \frac{\ln(0.5)}{19}\approx -0.03648$

Now, if the substance to decay to 78% of its original amount,

Then $A=78\% \text{ of }A_0 =\frac{78A_0}{100}=0.78 A_0$

$0.78 A_0=A_0 e^{-0.03648t}$

$0.78 = e^{-0.03648t}$

Again taking ln both sides,

$\ln(0.78) = -0.03648t$

$-0.24846=-0.03648t$

$\implies t = \frac{0.24846}{0.03648}=6.81085\approx 7$

Hence, approximately the substance would be 78% of its initial value after 7 years.

5 0

2 years ago