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11 months ago

A radioactive material has a half life of 10 years. What is the fraction of the initial isotope is left after 60 years

Mathematics

2 answers:

Alinara [238K]11 months ago

7 0

Answer:

1/64

Step-by-step explanation:

10 = ½

20 = ¼

30 = ⅛

40 = 1/16

50 = 1/32

60 = 1/64

Vanyuwa [196]11 months ago

3 0

Answer:

0.015625 left.

Step-by-step explanation:

If we start with say 100g of the material, in 10 years there is 100 * 1/2 =

50g left ( = 0.5 left).

In 20 years there is 100*(1/2)^2 = 25 g left.

In n*10 years there is 100(1/2)^2 g

So in 60 years there is 100(1/2)^6 = 1.5625 g left

This is 1.5625 %

= 0.015625 of the origin material left.

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melamori03 [73]

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From the question, we are to determine the product of the given expressions

The given expressions are

$\frac{x^{2} - 16}{2x+8}\times \frac{x^{3}-2x^{2} +x}{x^{2} +3x-4}$

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OR
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Hence, the product of the given expression is $\frac{x(x-1)(x-4)}{2(x+4)}$ OR $\frac{x(x^{2} -5x+4)}{2x+8}$ OR $\frac{x^{3} -5x^{2} +4x}{2x+8}$

Learn more on Simplifying fractions here: brainly.com/question/27038950

Here is the complete and correct question:

What is the product? StartFraction x squared minus 16 Over 2 x + 8 EndFraction times StartFraction x cubed minus 2 x squared + x Over x squared + 3 x minus 4 EndFraction

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