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FinnZ [79.3K]
3 years ago
11

Suppose that the halflife of a substance is 250 years . If there were ir 100 the substance , (a) give an exponential model for t

he situation, and (b) how much wil years?
can anyone answer this please:(
Mathematics
1 answer:
IgorLugansk [536]3 years ago
5 0
After 500 years one would have N nuclei of the original substance.

N = 100 * exp ( -ln(4) ) = 25

LOL.

I guess it is exp( - lamda * t)

Lamda, the decay constant is the ln(2) over the half life.

So we get exp( -ln(2)/250 * 500) or exp ( -ln(4) )
690 viewsView 1 Upvoter
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Factor the expression using the two different techniques listed for Parts 1(a) and 1(b).
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Answer:

9a^4 b^10(2 +3a^6 b^5)  (2 -3a^6 b^5)

see work below

Step-by-step explanation:

36a^4b^10 - 81a^16b^20

A)  find the GCF

36a^4b^10 = 4*9 a^4b^10  = 2*2*3*3 a*a*a*a*b*b*b*b*b*b*b*b*b*b

81a^16b^20= 9*9a^16b^20= 3*3*3*3* a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a                *b*b*b*b*b*b*b*b*b*b*b*b*b*b*b*b*b*b*b*b

The terms that appear in both terms is the GCF.  The terms that remain are inside the parentheses.

The GCF is 3*3 a*a*a*a*b*b*b*b*b*b*b*b*b*b

36a^4b^10 - 81a^16b^20 = 3*3 a*a*a*a*b*b*b*b*b*b*b*b*b*b *

( 2*2 - 3*3a*a*a*a*a*a*a*a*a*a*a*a b*b*b*b*b*b*b*b*b*b)

Combining like terms

36a^4b^10 - 81a^16b^20 = 9a^4b^10(4-9a^12b^10)

The expression inside the parenthesis can be factored using the difference of squares

let x^2 =4   x =2  

y^2 = 9a^12 b^10   y = 3a^6b^5

(x^2 -y^2) = (x+y)(x-y)

9a^4b^10(4-9a^12b^10) = 9a^4b^10 ( 2+3a^6b^5) ( 2-3a^6b^5)

b) difference of squares  a^2 – b^2 = (a + b)(a – b)

let a^2 = 36a^4b^10

so a = 6a^2b^5

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b = 9a^8 b^10

a^2 – b^2 = (a + b)(a – b)

36a^4b^10 - 81a^16b^20 = (6a^2b^5 +9a^8 b^10) (6a^2b^5 -9a^8 b^10)

We can factor a 3 a^2 b^5 out of the first term

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Multiply the terms outside the parentheses together

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