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Elina [12.6K]
3 years ago
14

Given that cos 0 = and sin 0 = - 15 What is the value of tan0?

Mathematics
1 answer:
SSSSS [86.1K]3 years ago
3 0

Step-by-step explanation:

\because \:  \cos \theta = \sin \theta =  - 15 \\  \\  \therefore \:  \:  \: \cos \theta = - 15 \:  \:  \\  \:  \:  \:  \:  \:  \:  \: and \:  \:  \\  \:  \:  \:  \:  \:  \:  \:  \:  \sin \theta =  - 15  \\ \therefore \:  \frac{\sin \theta }{\cos \theta}  =  \frac{ - 15}{ - 15} \\ \therefore \:   \huge \purple{ \boxed{\tan \theta  = 1}}

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10 men and 5 women are meeting in a conference room. Four people are chosen at random from the 15 to form a committee. (a) What
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Answer:

Step-by-step explanation:

Given

There are 10 men and 5 women in a conference room

total People=15

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The half-life of cesium-137 is 30 years. Suppose we have a 180-mg sample. (a) Find the mass that remains after t years. (b) How
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Answer:

a) Q(t) = 180e^{-0.023t}

b) 11.4mg of cesium-137 remains after 120 years.

c) 225.8 years.

Step-by-step explanation:

The following equation is used to calculate the amount of cesium-137:

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

In which Q(t) is the amount after t years, Q(0) is the initial amount, and r is the rate at which the amount decreses.

(a) Find the mass that remains after t years.

The half-life of cesium-137 is 30 years.

This means that Q(30) = 0.5Q(0). We apply this information to the equation to find the value of r.

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

0.5Q(0) = Q(0)e^{-30r}

e^{-30r} = 0.5

Applying ln to both sides of the equality.

\ln{e^{-30r}} = \ln{0.5}

-30r = \ln{0.5}

r = \frac{\ln{0.5}}{-30}

r = 0.023

So

Q(t) = Q(0)e^{-0.023t}

180-mg sample, so Q(0) = 180

Q(t) = 180e^{-0.023t}

(b) How much of the sample remains after 120 years?

This is Q(120).

Q(t) = 180e^{-0.023t}

Q(120) = 180e^{-0.023*120}

Q(120) = 11.4

11.4mg of cesium-137 remains after 120 years.

(c) After how long will only 1 mg remain?

This is t when Q(t) = 1. So

Q(t) = 180e^{-0.023t}

1 = 180e^{-0.023t}

e^{-0.023t} = \frac{1}{180}

e^{-0.023t} = 0.00556

Applying ln to both sides

\ln{e^{-0.023t}} = \ln{0.00556}

-0.023t = \ln{0.00556}

t = \frac{\ln{0.00556}}{-0.023}

t = 225.8

225.8 years.

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