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rosijanka [135]
3 years ago
7

Find the range of f(x) = -x2 + 1 for the domain {-2, 0, 2}

Mathematics
1 answer:
wariber [46]3 years ago
6 0
Range is just input the domain

so find f(-2), f(0) and f(2)

f(-2)=-(-2)²+1=-4+1=-3
f(0)=-(0)²+1=0+1=1
f(2)=-(2)²+1=-4+1=-3

range={-3,1}
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Pregnant women metabolize some drugs at a slower rate than the rest of the population. The half-life of caffeine is about 4 hour
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Answer:

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The husband will have 16.35 mg of caffeine in his body at 7 pm.

Woman:

The pregnant woman will have 51.33 mg of caffeine in her body at 7 pm.

Step-by-step explanation:

The amount of caffeine in the body can be modeled by the following equation:

C(t) = C(0)e^{rt}

In which C(t) is the amount of caffeine t hours after 8 am, C(0) is how much coffee they took and r is the rate the the amount of caffeine decreases in their bodies.

110 mg of caffeine at 8 am,

So C(0) = 110

Husband

Half life of 4 hours. So

C(4) = 0.5C(0) = 0.5*110 = 55

C(t) = C(0)e^{rt}

55 = 110e^{4r}

e^{4r} = 0.5

Applying ln to both sides

\ln{e^{4r}} = \ln{0.5}

4r = \ln{0.5}

r = \frac{\ln{0.5}}{4}

r = -0.1733

So for the husband

C(t) = 110e^{-0.1733t}

At 7 pm

7 pm is 11 hours after 8 am, so this is C(11)

C(t) = 110e^{-0.1733t}

C(11) = 110e^{-0.1733*11} = 16.35

The husband will have 16.35 mg of caffeine in his body at 7 pm.

Pregnant woman

Half life of 10 hours. So

C(10) = 0.5C(0) = 0.5*110 = 55

C(t) = C(0)e^{rt}

55 = 110e^{10}

e^{10r} = 0.5

Applying ln to both sides

\ln{e^{10r}} = \ln{0.5}

10r = \ln{0.5}

r = \frac{\ln{0.5}}{10}

r = -0.0693

At 7 pm

7 pm is 11 hours after 8 am, so this is C(11)

C(t) = 110e^{-0.0693t}

C(11) = 110e^{-0.0693*11} = 51.33

The pregnant woman will have 51.33 mg of caffeine in her body at 7 pm.

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