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HACTEHA [7]
3 years ago
15

5 grams of chucknorrisidium created, decays at a half-life of 16,000,000 Need 1 gram for experiments. How long before the is 1 g

ram remaining
Mathematics
1 answer:
devlian [24]3 years ago
8 0

Answer:

the time taken for the element to remain 1 gram is 38,400,000 unit.

Step-by-step explanation:

Given;

initial mass, m = 5 g

half life, t = 16,000,000 unit

The time taken for the element to remain 1 gram is calculated as;

          mass remaining                                      time

             5 g                                                        0

             2.5 g                                                    16,000,000

             1.25 g                                                   32,000,000

            0.625 g                                                 48,000,000

Interpolate to obtain the time at 1 gram;

1.25 g ----------------------------------   32,000,000

1 g --------------------------------------------- x

0.625 g --------------------------------------- 48,000,000

\frac{1.25 - 1}{1.25 - 0.625} = \frac{32,000,000 - x}{32,000,000 - 48,000,000} \\\\\frac{0.25}{0.625} = \frac{32,000,000 - x}{-16,000,000} \\\\-4,000,000= 20,000,000 - 0.625x\\\\0.625x = 20,000,000 + 4,000,000\\\\0.625x =  24,000,000\\\\x = \frac{24,000,000}{0.625} \\\\x = 38,400,000

Therefore, the time taken for the element to remain 1 gram is 38,400,000 unit.

         

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5 0
4 years ago
Which is the inverse of the function a(d)=5d-3? And use the definition of inverse functions to prove a(d) and a-1(d) are inverse
Drupady [299]

Answer:

a'(d) = \frac{d}{5} + \frac{3}{5}

a(a'(d)) = a'(a(d)) = d

Step-by-step explanation:

Given

a(d) = 5d - 3

Solving (a): Write as inverse function

a(d) = 5d - 3

Represent a(d) as y

y = 5d - 3

Swap positions of d and y

d = 5y - 3

Make y the subject

5y = d + 3

y = \frac{d}{5} + \frac{3}{5}

Replace y with a'(d)

a'(d) = \frac{d}{5} + \frac{3}{5}

Prove that a(d) and a'(d) are inverse functions

a'(d) = \frac{d}{5} + \frac{3}{5} and a(d) = 5d - 3

To do this, we prove that:

a(a'(d)) = a'(a(d)) = d

Solving for a(a'(d))

a(a'(d))  = a(\frac{d}{5} + \frac{3}{5})

Substitute \frac{d}{5} + \frac{3}{5} for d in  a(d) = 5d - 3

a(a'(d))  = 5(\frac{d}{5} + \frac{3}{5}) - 3

a(a'(d))  = \frac{5d}{5} + \frac{15}{5} - 3

a(a'(d))  = d + 3 - 3

a(a'(d))  = d

Solving for: a'(a(d))

a'(a(d)) = a'(5d - 3)

Substitute 5d - 3 for d in a'(d) = \frac{d}{5} + \frac{3}{5}

a'(a(d)) = \frac{5d - 3}{5} + \frac{3}{5}

Add fractions

a'(a(d)) = \frac{5d - 3+3}{5}

a'(a(d)) = \frac{5d}{5}

a'(a(d)) = d

Hence:

a(a'(d)) = a'(a(d)) = d

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andreev551 [17]

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