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topjm [15]
3 years ago
7

What is the distance between the points (–3, 4) and (–7, 4)?

Mathematics
2 answers:
Liono4ka [1.6K]3 years ago
6 0
10 I hope this helps bye
Elodia [21]3 years ago
3 0
Distance between (x1,y1) and (x2,y2) is
D= \sqrt{(x2-x1)^2+(y2-y1)^2}

given
(-3,4) and (-7,4)

D= \sqrt{(-7-(-3))^2+(4-4)^2}
D= \sqrt{(-7+3)^2+(0)^2}
D= \sqrt{(-4)^2+0}
D=√16
D=4

distance is 4
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Initially 100 milligrams of a radioactive substance was present. After 6 hours the mass had decreased by 3%. If the rate of deca
Hitman42 [59]

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The half-life of the radioactive substance is 135.9 hours.

Step-by-step explanation:

The rate of decay is proportional to the amount of the substance present at time t

This means that the amount of the substance can be modeled by the following differential equation:

\frac{dQ}{dt} = -rt

Which has the following solution:

Q(t) = Q(0)e^{-rt}

In which Q(t) is the amount after t hours, Q(0) is the initial amount and r is the decay rate.

After 6 hours the mass had decreased by 3%.

This means that Q(6) = (1-0.03)Q(0) = 0.97Q(0). We use this to find r.

Q(t) = Q(0)e^{-rt}

0.97Q(0) = Q(0)e^{-6r}

e^{-6r} = 0.97

\ln{e^{-6r}} = \ln{0.97}

-6r = \ln{0.97}

r = -\frac{\ln{0.97}}{6}

r = 0.0051

So

Q(t) = Q(0)e^{-0.0051t}

Determine the half-life of the radioactive substance.

This is t for which Q(t) = 0.5Q(0). So

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0.5Q(0) = Q(0)e^{-0.0051t}

e^{-0.0051t} = 0.5

\ln{e^{-0.0051t}} = \ln{0.5}

-0.0051t = \ln{0.5}

t = -\frac{\ln{0.5}}{0.0051}

t = 135.9

The half-life of the radioactive substance is 135.9 hours.

6 0
3 years ago
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