Find the length of AC

First,

tan(θ) = sin(θ) / cos(θ)

and given that 90° < θ < 180°, meaning θ lies in the second quadrant, we know that cos(θ) < 0. (We also then know the sign of sin(θ), but that won't be important.)

Dividing each part of the inequality by 2 tells us that 45° < θ/2 < 90°, so the half-angle falls in the first quadrant, which means both cos(θ/2) > 0 and sin(θ/2) > 0.

Now recall the half-angle identities,

cos²(θ/2) = (1 + cos(θ)) / 2

sin²(θ/2) = (1 - cos(θ)) / 2

and taking the positive square roots, we have

cos(θ/2) = √[(1 + cos(θ)) / 2]

sin(θ/2) = √[(1 - cos(θ)) / 2]

Then

tan(θ/2) = sin(θ/2) / cos(θ/2) = √[(1 - cos(θ)) / (1 + cos(θ))]

Notice how we don't need sin(θ) ?

Now, recall the Pythagorean identity:

cos²(θ) + sin²(θ) = 1

Dividing both sides by cos²(θ) gives

1 + tan²(θ) = 1/cos²(θ)

We know cos(θ) is negative, so solve for cos²(θ) and take the negative square root.

cos²(θ) = 1/(1 + tan²(θ))

cos(θ) = - 1/√[1 + tan²(θ)]

Plug in tan(θ) = - 12/5 and solve for cos(θ) :

cos(θ) = - 1/√[1 + (-12/5)²] = - 5/13

Finally, solve for sin(θ/2) and tan(θ/2) :

sin(θ/2) = √[(1 - (- 5/13)) / 2] = 3/√(13)

tan(θ/2) = √[(1 - (- 5/13)) / (1 + (- 5/13))] = 3/2

Answer:

$3.50

Step-by-step explanation:

14 is double 7

So the price should be double

1.75 x 2 = 3.5

(Assuming there's no base fee and the charge is purely for minutes)

Answer:

There is nothing here?

Step-by-step explanation:

Answer:

99.7%

Step-by-step explanation:

99.7% of boys fall between 61.5 and 76.5

The percentage of the boys that are between 61.5 and 76.5 inches tall is 99.73%

The given parameters are:

Start by calculating the z score for x = 61.5 and 76.5 using:

[tex]z = \frac{x - \mu}{\sigma}[/tex]

So, we have:

[tex]z_1 = \frac{61.5 - 69}{2.5} = -3[/tex]

[tex]z_2 = \frac{76.5 - 69}{2.5} = 3[/tex]

The percentage is then represented as:

Percentage = P(-3 < x < 3)

Using the z table of probabilities, we have:

Percentage = 0.9973

Express as percentage

Percentage = 99.73%

Hence, 99.73% of the boys are between 61.5 and 76.5 inches tall

Read more about normal distribution at:

https://brainly.com/question/4079902

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