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3 years ago

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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Answer:

Step-by-step explanation:

Given

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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.

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

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

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Applying ln to both sides of the equality.

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

$-30r = \ln{0.5}$

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

$r = 0.023$

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

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).

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

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$Q(120) = 11.4$

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

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225.8 years.

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