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Vedmedyk [2.9K]
3 years ago
10

Two linear are parallel if and only if their slopes are

Mathematics
1 answer:
grin007 [14]3 years ago
8 0
Their slopes must be the same, but with different y-intercepts.

Here is an example: 2x + 1 and 2x + 4 are parallel lines.
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Find the real zeros of the trigonometric function on the interval 0 ≤ x < 2π
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4[cos(x)]^2 = 3

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That happens in the first and fourth quadrants, for the angles 30 degrees and 330 degrees.

Answer: x = 30 degrees and x = 330 degrees
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Two banks are running promotions on savings accounts in anticipation of higher interest rates in the future.
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<h2>First Federal Bank is best</h2>

Step-by-step explanation:

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The sample space, S, for flipping a fair coin three times is shown
Ksenya-84 [330]

Answer: \frac{7}{8}

Step-by-step explanation:

Given

Sample space is ={HHH,THH,HTH,HHT,TTH,THT,HTT,TTT}

There are 8 element in sample space which indicates the outcome of tossing a coin three times

Probability of obtaining at least one head=1-P(\text{0 head})

P(\text{0 head})=P(\text{All tails})

P=1-\frac{1}{8}

P=\frac{7}{8}

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3 years ago
Eva reads 12 pages per hour. In all, how many hours reading will Eva have to do this week in order to have read a total of 48 pa
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Since eva is a slow reader pacing at 12 pages an hour, she would have to read for 4 hours over the week in order to read 48 pages total.

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3 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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