# How do you find x if the point (x,-3) is on the terminal side of theta and #sintheta=-3/5#?

According to sine's definition, we have

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To find ( x ), where ( (\text{x}, -3) ) is on the terminal side of ( \theta ) and ( \sin(\theta) = -\frac{3}{5} ), we can use the Pythagorean identity for sine and cosine. Since ( \sin(\theta) = \frac{y}{r} ) in a polar coordinate system, and in this case ( y = -3 ), we can find ( r ) using the Pythagorean theorem (( r^2 = x^2 + y^2 )). Then, we can solve for ( x ) using the equation ( x = r \cdot \cos(\theta) ).

First, calculate ( r ): [ r^2 = x^2 + (-3)^2 ] [ r^2 = x^2 + 9 ] Since ( \sin(\theta) = -\frac{3}{5} ), and ( \sin(\theta) = \frac{y}{r} ), we have: [ -\frac{3}{5} = \frac{-3}{r} ] [ r = 5 ]

Now that we have ( r ), we can find ( x ): [ x = r \cdot \cos(\theta) ] [ x = 5 \cdot \cos(\theta) ]

Now, to find ( \cos(\theta) ), use the fact that ( \sin^2(\theta) + \cos^2(\theta) = 1 ): [ \cos^2(\theta) = 1 - \sin^2(\theta) ] [ \cos^2(\theta) = 1 - \left(-\frac{3}{5}\right)^2 ] [ \cos^2(\theta) = 1 - \frac{9}{25} ] [ \cos^2(\theta) = \frac{16}{25} ] [ \cos(\theta) = \pm \frac{4}{5} ]

Since ( x ) is on the left side (negative ( x ) axis), we choose the negative value for ( \cos(\theta) ): [ x = 5 \cdot \left(-\frac{4}{5}\right) ] [ x = -4 ]

Therefore, ( x = -4 ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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