To evaluate a function<span>, I </span>do<span> just what I </span>did<span> above: I plug in the given value for x. Here, I am supposed to </span>evaluate<span> at the value x = –3. The notation is different, but "f(–3)" </span>means<span> exactly the same thing as "</span>evaluate<span> at x = –3"! Note how I used parentheses.</span>
Answer:
x= -32
Step-by-step explanation:
The answer is x= -32
Answer:
The independent variable is number of minutes and the dependent variable is number of miles. Constant of proportionality is 1.2 because 9.6/8=1.2 and 18/15=1.2
Step-by-step explanation:
Answer:The difference between x and y is 1. In other words, x minus y equals 1 and can be written as equation B: · x - y = 1. Now solve equation B for x
Step-by-step explanation:
Recall the Pythagorean identity,
![\sin^2(\theta) + \cos^2(\theta) = 1](https://tex.z-dn.net/?f=%5Csin%5E2%28%5Ctheta%29%20%2B%20%5Ccos%5E2%28%5Ctheta%29%20%3D%201)
Since
belongs to Q3, we know both
and
are negative. Then
![\cos(\theta) = -\sqrt{1 - \sin^2(\theta)} = -\dfrac7{25}](https://tex.z-dn.net/?f=%5Ccos%28%5Ctheta%29%20%3D%20-%5Csqrt%7B1%20-%20%5Csin%5E2%28%5Ctheta%29%7D%20%3D%20-%5Cdfrac7%7B25%7D)
Recall the half-angle identities for sine and cosine,
![\sin^2\left(\dfrac\theta2\right) = \dfrac{1 - \cos(\theta)}2](https://tex.z-dn.net/?f=%5Csin%5E2%5Cleft%28%5Cdfrac%5Ctheta2%5Cright%29%20%3D%20%5Cdfrac%7B1%20-%20%5Ccos%28%5Ctheta%29%7D2)
![\cos^2\left(\dfrac\theta2\right) = \dfrac{1 + \cos(\theta)}2](https://tex.z-dn.net/?f=%5Ccos%5E2%5Cleft%28%5Cdfrac%5Ctheta2%5Cright%29%20%3D%20%5Cdfrac%7B1%20%2B%20%5Ccos%28%5Ctheta%29%7D2)
Then by definition of tangent,
![\tan^2\left(\dfrac\theta2\right) = \dfrac{\sin^2\left(\frac\theta2\right)}{\cos^2\left(\frac\theta2\right)} = \dfrac{1 - \cos(\theta)}{1 + \cos(\theta)}](https://tex.z-dn.net/?f=%5Ctan%5E2%5Cleft%28%5Cdfrac%5Ctheta2%5Cright%29%20%3D%20%5Cdfrac%7B%5Csin%5E2%5Cleft%28%5Cfrac%5Ctheta2%5Cright%29%7D%7B%5Ccos%5E2%5Cleft%28%5Cfrac%5Ctheta2%5Cright%29%7D%20%3D%20%5Cdfrac%7B1%20-%20%5Ccos%28%5Ctheta%29%7D%7B1%20%2B%20%5Ccos%28%5Ctheta%29%7D)
belonging to Q3 means
, or
, so that the half-angle belongs to Q2. Then
is positive and
is negative, so
is negative.
It follows that
![\tan\left(\dfrac\theta2\right) = -\sqrt{\dfrac{1 - \cos(\theta)}{1 + \cos(\theta)}} = \boxed{-\dfrac43}](https://tex.z-dn.net/?f=%5Ctan%5Cleft%28%5Cdfrac%5Ctheta2%5Cright%29%20%3D%20-%5Csqrt%7B%5Cdfrac%7B1%20-%20%5Ccos%28%5Ctheta%29%7D%7B1%20%2B%20%5Ccos%28%5Ctheta%29%7D%7D%20%3D%20%5Cboxed%7B-%5Cdfrac43%7D)