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3 years ago

20 POINTS PLEASE HELP

Mathematics

2 answers:

Ad libitum [116K]3 years ago

8 0

Answer:

17y

Step-by-step explanation:

8y+6y+3y= 17y

Andrei [34K]3 years ago

3 0

17y I think
8y+6y+3y=17y

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Solve for x : 6x+ 1/4 (4x+8)> 12

loris [4]

Answer:

x > 10/7 or x > 1 3/7

Step-by-step explanation:

First simplify the left side of the inequality.

6x+ 1/4 (4x+8)> 12

6x+x+2>12

7x+2>12 Next, using the property of inequality, subtract two from both sides.

7x>10 Now divide by 7 to solve for x.

x>10/7 or x> 1 3/7

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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3 years ago

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Step-by-step explanation:

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3 years ago

Suppose after 2500 years an initial amount of 1000 grams of a radioactive substance has decayed to 75 grams. What is the half-li

krok68 [10]

Answer:

The correct answer is:

Between 600 and 700 years (B)

Step-by-step explanation:

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:

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

Therefore the half-life is between 600 and 700 years

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3 years ago