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maria [59]
3 years ago
8

Evaluate the expresion for x=6. -2x + 1/3(x2 + 6) - 36/6

Mathematics
2 answers:
yulyashka [42]3 years ago
7 0

Answer:

-12

Step-by-step explanation:

-2(6)+1/3(12+6)-6

-12+1/3(18)-6

-12+6-6

-12

Korvikt [17]3 years ago
3 0

Answer:

x=6/7            hope i did it right- im so sorry if thats wrong

Step-by-step explanation:

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Determine if the given lines are parallel, perpendicular or neither.<br> 2x + 7y = 4<br> y = 3x + 5
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3 years ago
Use the following scenario to answer the question below:
Aleks [24]

Answer:

9 hr * 60 min/hr = 540 min. 540 min * 60 sec/min = 32,400 sec. Use the method described by panic mode: round the number down until there is only one non-zero digit left. Here, 32,400 rounds down to 30,000. Now count the number of zeroes; the result is the order of magnitude: 4. (You see this in scientific notation also: 30,000 = 3 × 10^4.)

5 0
2 years ago
Pregnant women metabolize some drugs at a slower rate than the rest of the population. The half-life of caffeine is about 4 hour
UkoKoshka [18]

Answer:

Husband:

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.

7 0
3 years ago
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