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3 years ago

Consider the points R and S in the figure. How many different lines pass through both R and S? Explain.

Mathematics

1 answer:

OLga [1]3 years ago

3 0

Answer:

Step-by-step explanation:

Two distinct points define exactly one line.

___

We haven't seen the figure. If R and S are the same point, an infinite number of lines will pass through R and S.

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What is −5/8÷9 1/0? please help

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Answer:

Step-by-step explanation:

the answer is 0 because it is 1/0

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4 years ago

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The following question find the value of the variables. If your answer is not an integer leave it in simplest radical form

Margarita [4]

Answer:

option 4

Step-by-step explanation:

using the sine/ cosine ratios in the right triangle and the exact values

cos30° = $\frac{\sqrt{3} }{2}$ , sin30° = $\frac{1}{2}$ , then

cos30° = $\frac{adjacent}{hypotenuse}$ = $\frac{x}{20\sqrt{3} }$ = $\frac{\sqrt{3} }{2}$ ( cross- multiply )

2x = 20 $\sqrt{3}$ × $\sqrt{3}$ = 20 × 3 = 60 ( divide both sides by 2 )

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and

sin30° = $\frac{opposite}{hypotenuse}$ = $\frac{y}{20\sqrt{3} }$ = $\frac{1}{2}$ ( cross- multiply )

2y = 20 $\sqrt{3}$ ( divide both sides by 2 )

y = 10 $\sqrt{3}$

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2 years ago

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C I believe that it is C also I have bad period cramps send chocolate

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3 years ago

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I need help plz help

mihalych1998 [28]

Answer:

132=5x+7 because opposisie angle are equal

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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.

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4 years ago