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madam [21]
3 years ago
6

Find all angles θ between 0° and 180° satisfying the given equation. round your answer to one decimal place. (enter your answers

as a comma-separated list.) tan(θ) = −21

Mathematics
1 answer:
scoray [572]3 years ago
6 0

Alright, lets get started.

tan Ф = -21

Using inverse function,

Ф = tan^{-1} (-21)

Using calculator, we could find ta inverse value

Ф = - 87.3°

But as per given question, angle must lie between 0 and 180 degrees.

Means angle should present in first or second quadrant.

As tan value is given negative (-21), means angle will be in second quadrant.

If we add 180 degrees in our angle, the angle will lie in second quadrant. (refer the diagram)

Ф = - 87.27 + 180 = 92.73°

Ф = 92.73° : Answer

Hope it will help :)

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Answer:

a. y=6(1.7472)^x

b. y=6e^{0.558t}

c.13.3 months

Step-by-step explanation:

a.-Given the first term  at t_0 is 6 and the second term at t_3 is 32.

-Let's take rabbit population as a function of time to be

y=ab^x

where y is the population at time x and a the initial population at t_0\\

#We substitute our values to calculate the value of the constant b:

y_x=ab^x\\\\y_3=ab^3\\\\32=6b^3\\\\b=1.472

#Replace b in the population function:

y=ab^x, b=1.7472,a=6\\\\\therefore y=6(1.7472)^x

Hence, the regression for the rabbit population as a function of time x is y=6(1.7472)^x

b. The exponential function in terms of base e is usually expressed as:

A=A_0e^{kt}

Where:

A_0-is the initial population at t_o

A-is the population at time t.

k-is the  exponential growth constant.

e- the exponent

Our function in terms of base exponent is rewritten as:

y=A_0e^{kt}

#Substitute with actual figures to solve for t:

y=A_0e^{kt}, y=32, xt=3, A_0=6\\\\32=6e^{3k}\\\\3k=In (32/6)\\\\k=0.5580

Hence, the regression equation in terms of base e is y=6e^{0.558t}

c. We substitute y with any number higher than 10,000 to estimate the time for the rabbits to exceed 10,000.

-We know that y=6e^{0.558t}.

Therefore we calculate t as(take y=10001):

y=6e^{0.558t}, y=10001\\\\10001=6e^{0.558t}\\\\1666.8333=e^{0.558t}\\\\0.558t= In 1666.8333\\\\t=13.2951

Hence, it takes approximately 13.3 months for the population to exceed 10000

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