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stiv31 [10]
3 years ago
14

Find all angles, 0≤θ<360, that satisfy the equation below, to the nearest 10th of a degree. 4tan^2 θ+tanθ=−4tanθ−1

Mathematics
1 answer:
Serga [27]3 years ago
3 0

Answer:

135 and 315degrees

Step-by-step explanation:

Given the expression

4tan²θ+tanθ=−4tanθ−1

Equate to zero

4tan²θ+tanθ+4tanθ+1 = 0

4tan²θ+5tanθ+1 = 0

Let x = tanθ

4x²+5x+1 = 0

Factorize;

4x²+4x+x+1 = 0

4x(x+1)+1(x+1) = 0

4x+1 = 0 and x+1 = 0

4x = -1 and x = -1

x = -1/4 and -1

Since x = tanθ

-1 =tanθ

θ = arctan(-1)

θ = -45degrees

Since tan is negative in the 2nd and fourth quadrant;

In the second quadrant

θ = 180 - 45 = 135degrees

In the fourth quadrant;

θ = 360 - 45 = 315degrees

Hence the required angles are 135 and 315degrees

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Step-by-step explanation:

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Simplify 15 to the 18th power over 15 to the 3rd power.
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The table compares x, the number of minutes of television Sam watched each day, to y, the number of minutes she spent exercising
Ahat [919]

For this case we have the following variables:

x: the number of minutes of television Sam watched each day

y: the number of minutes she spent exercising

Then, we have the following function that models the problem:

y = -0.78x + 95

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Substituting in the given equation we have:

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Answer:

the best prediction for the number of minutes of exercising Sam will do if she spends 30 minutes watching television that day is:

72 minutes

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A line passes through the point (6, -2) and has a slope of 4/3 Write an equation in point-slope form for this line.​
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Answer:

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5 0
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You are the coordinator for a program that is going to take place at night in a rectangular amphitheater in the mountains. You w
White raven [17]

The equation to model the situation is \mathbf{y = \dfrac{k}{x^2}}. The constant for the variation is 2250.

<h3>What is the intensity of light?</h3>

The intensity of light from a lantern varies inversely to the square of the distance from the lantern.

From the given information:

  • Let y be the intensity of light, and
  • x be the distance from the lantern

Then:

\mathbf{y \alpha \dfrac{1}{x^2} }

\mathbf{y = \dfrac{k}{x^2} }   here, k = constant.

2.

If y = 90 W/m² when the distance x = 5m

Then:

\mathbf{90 = \dfrac{k}{(5)^2}}

k = 90 × 25

k = 2250

c.

The equation to model the situation by using the constant variation is:

\mathbf{y = \dfrac{2250}{x^2}}

d.

If the light intensity y = 40, then x is determined as:

\mathbf{40 = \dfrac{2250}{x^2}}

\mathbf{x = \sqrt{\dfrac{2250}{40}}}

x = 7.5 m

e.

The light is needed in (225 × 1000)m = 225000 km of illumination.

f.

The lantern required for the new light estimation is:

y = 2250/225000

y = 0.01 intensity

Therefore, we can conclude that to get an intensity of 1 W/m², we need to put 100 lanterns.

Learn more about intensity of light here:

brainly.com/question/19791748

#SPJ1

4 0
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