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Artyom0805 [142]
3 years ago
15

Which relation is a function of x?

Mathematics
2 answers:
Monica [59]3 years ago
5 0

Answer:

Table D

Step-by-step explanation:

I just took the quiz on edge

USPshnik [31]3 years ago
3 0

Answer:

The fourth function

Step-by-step explanation:

You can never have repeating x-values. You can have repeating y-values though. Whenever determining whether something is a function, just look at whether it has repeating x-values or not. If it does, it is not a function. If it doesn't, it is a function.

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For 0 ≤ ϴ < 2π, how many solutions are there to tan(StartFraction theta Over 2 EndFraction) = sin(ϴ)? Note: Do not include va
Black_prince [1.1K]

Answer:

3 solutions:

\theta={0, \frac{\pi}{2}, \frac{3\pi}{2}}

Step-by-step explanation:

So, first of all, we need to figure the angles that cannot be included in our answers out. The only function in the equation that isn't defined for some angles is tan(\frac{\theta}{2}) so let's focus on that part of the equation first.

We know that:

tan(\frac{\theta}{2})=\frac{sin(\frac{\theta}{2})}{cos(\frac{\theta}{2})}

therefore:

cos(\frac{\theta}{2})\neq0

so we need to find the angles that will make the cos function equal to zero. So we get:

cos(\frac{\theta}{2})=0

\frac{\theta}{2}=cos^{-1}(0)

\frac{\theta}{2}=\frac{\pi}{2}+\pi n

or

\theta=\pi+2\pi n

we can now start plugging values in for n:

\theta=\pi+2\pi (0)=\pi

if we plugged any value greater than 0, we would end up with an angle that is greater than 2\pi so,  that's the only angle we cannot include in our answer set, so:

\theta\neq \pi

having said this, we can now start solving the equation:

tan(\frac{\theta}{2})=sin(\theta)

we can start solving this equation by using the half angle formula, such a formula tells us the following:

tan(\frac{\theta}{2})=\frac{1-cos(\theta)}{sin(\theta)}

so we can substitute it into our equation:

\frac{1-cos(\theta)}{sin(\theta)}=sin(\theta)

we can now multiply both sides of the equation by sin(\theta)

so we get:

1-cos(\theta)=sin^{2}(\theta)

we can use the pythagorean identity to rewrite sin^{2}(\theta) in terms of cos:

sin^{2}(\theta)=1-cos^{2}(\theta)

so we get:

1-cos(\theta)=1-cos^{2}(\theta)

we can subtract a 1 from both sides of the equation so we end up with:

-cos(\theta)=-cos^{2}(\theta)

and we can now add cos^{2}(\theta)

to both sides of the equation so we get:

cos^{2}(\theta)-cos(\theta)=0

and we can solve this equation by factoring. We can factor cos(\theta) to get:

cos(\theta)(cos(\theta)-1)=0

and we can use the zero product property to solve this, so we get two equations:

Equation 1:

cos(\theta)=0

\theta=cos^{-1}(0)

\theta={\frac{\pi}{2}, \frac{3\pi}{2}}

Equation 2:

cos(\theta)-1=0

we add a 1 to both sides of the equation so we get:

cos(\theta)=1

\theta=cos^{-1}(1)

\theta=0

so we end up with three answers to this equation:

\theta={0, \frac{\pi}{2}, \frac{3\pi}{2}}

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