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.
Answer:
To make this easier, first draw or imagine a vertical line at x=2
Find where the vertical line and the graph intersects.
Draw or imagine that point of intersection.
Now, draw or imagine a horizontal line that goes through the point of intersection.
That horizontal line tells us the the value of the function by passing through that mark that says "-2"
The function's value at x=2 is -2.
Have an awesome day! :)
Step-by-step explanation:
Answer:
IT would be 13a to 17b
Step-by-step explanation:
I had this.
Answer:
f(1) = 6
Step-by-step explanation:
You just substitute 1 for x, 3+3 = 6
