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:
2. (5, -5)
3. (-3, 2)
4. (2, 4)
Step-by-step explanation:
Each vertex can be found using two simple steps.
1) The number being added or subtracted from x.
-The x coordinate of your vertex is the opposite of that number.
2) The number after the absolute value symbol.
-The y coordinate is that value.
cos4x = cos2x
We know that:
cos2x = 1-2cos^2 x
==> cos4x = 1-2cos^2 (2x)
Now substitute:
==> 1-2cos^2 (2x) = cos2x
==> 2cos^2 (2x) + cos2x - 1 = 0
Now factor:
==> (2cos2x -1)(cos2x + 1) = 0
==> 2cos2x -1 = 0 ==> cos2x =1/2 ==> 2x= pi/3
==> x1= pi/6 , 7pi/6
==> x1= pi/6 + 2npi
==> x2= 7pi/6 + 2npi
==> cos2x = -1 ==> 2x= pi ==> x3 = pi/2 + 2npi.
<span>==> x= { pi/6+2npi, 7pi/6+2npi, pi/2+2npi}</span>
Answer:
3h - 6
Step-by-step explanation:
To understand better,I have attached a picture to this answer
Answer:
12
Step-by-step explanation:
This is assuming that x2 means x to the power of 2 and that the x on the left is to represent multiplying. First, plug in all of the number to get 2*3+3^2-3.Using PEMDAS, you should solve the exponent to get 9. Then, multiply 2 by 3 to get 6. Add 6 to 9 to get 15. Subtract 2 to 15 to get 12.
TL;DR
2*3+3^2-3
6+9-3
15-3
12
