Problem: (5x^4 – 3x^3 + 6x) – (3x^3 + 11x^2 – 8x) Answer: 5x^4-6x^3 -11x^2 +14x

1 answer:

8 0

Combine the like terms by adding the coefficients,
->5x^4 can't add anything so leave it there,
->-3x^3 can add -3x^3 which gives you -6x^3,
->We can't do anything with -11x^2 so leave it there as well,
->6x add to -(-8x) gives you 14x,
You will get for your final answer is 5x^4-6x^3-11x^2+14x.

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At a constant decay rate, the half-life of a radioactive substance is the time taken for the substance to decay to half of its original mass. The formula for radioactive exponential decay is given by:

$A(t) = A_0 e^{(kt)}\\where:\\A(t) = Amount\ left\ at\ time\ (t) = 75\ grams\\A_0 = initial\ amount = 1000\ grams\\k = decay\ constant\\t = time\ of\ decay = 2500\ years$

First, let us calculate the decay constant (k)

$75 = 1000 e^{(k2500)}\\dividing\ both\ sides\ by\ 1000\\0.075 = e^{(2500k)}\\taking\ natural\ logarithm\ of\ both\ sides\\In 0.075 = In (e^{2500k})\\In 0.075 = 2500k\\k = \frac{In0.075}{2500}\\ k = \frac{-2.5903}{2500} \\k = - 0.001036$

Next, let us calculate the half-life as follows:

$\frac{1}{2} A_0 = A_0 e^{(-0.001036t)}\\Dividing\ both\ sides\ by\ A_0\\ \frac{1}{2} = e^{-0.001036t}\\taking\ natural\ logarithm\ of\ both\ sides\\In(0.5) = In (e^{-0.001036t})\\-0.6931 = -0.001036t\\t = \frac{-0.6931}{-0.001036} \\t = 669.02 years\\\therefore t\frac{1}{2} \approx 669\ years$

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