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
This procedure can be done for every angle , so you don't have to work with acute angles only.
You transformed the function as follows:
All trasformations of the fashion 
translate the function horizontally, k units right if k is positive, k units left if k is negative.
In your case, k=6, so you translated the original function y=|x| 6 units to the right.
Answer:
Do you have options? The answer would be
g(x)=(x/3)^2
x = 1 btw
Because x = 1, if you have options then it'll most likely be g(x)=(1/3]^2
Step-by-step explanation:
Hope this helps
Price after off =80- 1/4 * 80 = 80-20 = $60
Coupon discount = 60*10/100 = $6
So, he have to pay, 60-6 = $54
Drawing this square and then drawing in the four radii from the center of the cirble to each of the vertices of the square results in the construction of four triangular areas whose hypotenuse is 3 sqrt(2). Draw this to verify this statement. Note that the height of each such triangular area is (3 sqrt(2))/2.
So now we have the base and height of one of the triangular sections.
The area of a triangle is A = (1/2) (base) (height). Subst. the values discussed above, A = (1/2) (3 sqrt(2) ) (3/2) sqrt(2). Show that this boils down to A = 9/2.
You could also use the fact that the area of a square is (length of one side)^2, and then take (1/4) of this area to obtain the area of ONE triangular section. Doing the problem this way, we get (1/4) (3 sqrt(2) )^2. Thus,
A = (1/4) (9 * 2) = (9/2). Same answer as before.
