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Elan Coil [88]
1 year ago
7

Helppppp with alll the problems asap!! Show your work

Mathematics
1 answer:
Nikitich [7]1 year ago
6 0

17) x+2x+1+3x=\boxed{6x+1}

18) 17-a+5a-1+4ab=\boxed{4ab+4a+16}

19) 2(3y-5+y+2)=\boxed{8y-6}

20) k^2 +k+k^2 -9k+23+k^2 +k+2k^2 -17=\boxed{5k^2 -7k+6}

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polkadot's strips

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polkadot's plain

Step-by-step explanation:

3

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Need help with this math lesson
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A train leaves the station at 9:00 pm traveling east at 36 miles per hour. A second train leaves the station at 10:00 pm traveli
GrogVix [38]
X - hours that passed after 10 pm

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4 0
3 years ago
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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

5 0
3 years ago
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