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Serjik [45]
4 years ago
14

Which equation represents a circle with the same center as the circle shown but with a radius of 2 units

Mathematics
2 answers:
makkiz [27]4 years ago
4 0

Answer:

B. (x – 4)2 + (y – 5)2 = 4

Step-by-step explanation:

Leona [35]4 years ago
3 0

Answer:

Option B.

Step-by-step explanation:

Consider the below figure attached with this question.

From the below figure it is clear that the center of the given circle is (4,5).

The standard form of a circle is

(x-h)^2+(y-k)^2=r^2

where, (h,k) is center of the circle and r is radius.

We need to find the equation of a circle which represents the same center as the circle shown but with a radius of 2 units.

Substitute h=4, k=5 and r=2 in the above equation.

(x-4)^2+(y-5)^2=(2)^2

(x-4)^2+(y-5)^2=4

The required equation is (x-4)^2+(y-5)^2=4.

Therefore, the correct option is B.

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3 years ago
Suppose an unknown radioactive substance produces 8000 counts per minute on a Geiger counter at a certain time, and only 500 cou
mariarad [96]

Answer:

The half-life of the radioactive substance is of 3.25 days.

Step-by-step explanation:

The amount of radioactive substance is proportional to the number of counts per minute:

This means that the amount is given by the following differential equation:

\frac{dQ}{dt} = -kQ

In which k is the decay rate.

The solution is:

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

In which Q(0) is the initial amount:

8000 counts per minute on a Geiger counter at a certain time

This means that Q(0) = 8000

500 counts per minute 13 days later.

This means that Q(13) = 500. We use this to find k.

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

500 = 8000e^{-13k}

e^{-13k} = \frac{500}{8000}

\ln{e^{-13k}} = \ln{\frac{500}{8000}}

-13k = \ln{\frac{500}{8000}}

k = -\frac{\ln{\frac{500}{8000}}}{13}

k = 0.2133

So

Q(t) = Q(0)e^{-0.2133t}

Determine the half-life of the radioactive substance.

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

Q(t) = Q(0)e^{-0.2133t}

0.5Q(0) = Q(0)e^{-0.2133t}

e^{-0.2133t} = 0.5

\ln{e^{-0.2133t}} = \ln{0.5}

-0.2133t = \ln{0.5}

t = -\frac{\ln{0.5}}{0.2133}

t = 3.25

The half-life of the radioactive substance is of 3.25 days.

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