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:
Step-by-step explanation:
simplify (5-3)
PEMDAS= parentheses- e -MULTIPLY- DIVIDE - ADD- SUBTRACT
2(5-3)2+62
(2*2)2+62
8+62
70
Answer:
$468
Step-by-step explanation:
B
Its because when you rearrange using this formula y=mx+c, where m is slope, if m is negative then that line has a negative slope
I think we can use the identity sin x/2 = sqrt [(1 - cos x) /2]
cos x - sqrt3 sqrt ( 1 - cos x) /sqrt2 = 1
cos x - sqrt(3/2) sqrt(1 - cos x) = 1
sqrt(3/2)(sqrt(1 - cos x) = cos x - 1 Squaring both sides:-
1.5 ( 1 - cos x) = cos^2 x - 2 cos x + 1
cos^2 x - 0.5 cos x - 0.5 = 0
cos x = 1 , -0.5
giving x = 0 , 2pi, 2pi/3, 4pi/3 ( for 0 =< x <= 2pi)
because of thw square roots some of these solutions may be extraneous so we should plug these into the original equations to see if they fit.
The last 2 results dont fit so the answer is x = 0 , 2pi Answer
