Every point in the unit circle is identified either by its coordinates
or by the angle it forms with the x-axis,
.
The trigonometric functions associate with every angle
and the correspondant
coordinates the two values
![\cos(\alpha)=x,\quad\sin(\alpha)=y](https://tex.z-dn.net/?f=%5Ccos%28%5Calpha%29%3Dx%2C%5Cquad%5Csin%28%5Calpha%29%3Dy)
This procedure can be done for every angle
, so you don't have to work with acute angles only.
Answer:
39.0953=39cm2
Step-by-step explanation:
Minor arc =70
Angle of arc=70
Then this sector =70/360=7/36 of the circle
Then area = 7/36x3.14x8x8=39cm2 approximatly
Square Root of 7
= 2.64575131106
Answer:
Step-by-step explanation:
10(5)=50
The function in vertex form is
![f(t) = (t+6)^{2} -54](https://tex.z-dn.net/?f=f%28t%29%20%3D%20%28t%2B6%29%5E%7B2%7D%20-54)
(refer to your other post I solved it there).
The general form of quadratic equations in vertex form is
![f(x) = a(x-h)^{2} +k](https://tex.z-dn.net/?f=f%28x%29%20%3D%20a%28x-h%29%5E%7B2%7D%20%2Bk)
, where (h, k) is the vertex of the parabola.
Here, a = 1, h = -6 and k = -54
Therefore, the vertex is (-6, -54) and it is a maximum because a = 1 is postive.