13) 9 over 4 × 4 over 3
9 × 4 over 4 × 3
36 over 12 = 3 over 1
Therefore, your answer would be 3 over 1
14) 7 over 5 × 20 over 9
7 × 20 over 5 × 9
140 over 45 = 3 and 1 over 9
Therefore, your answer would have to be: 3 and 1 over 9.
:)
There is a square and a half circle.
To find the radius of the circle: 18in - 12in = 6in
Area of a circle = (pi)r^2 = 3.14 (6in)^2 = 3.14 (36in^2) = 113.04in^2
This is an half circle, so we have to divide by 2 the result
56.52in^2 in what we’re looking for.
Area of the square = (12in)^2 = 144in^2
Total area = (144 + 56.52)in^2 = 200.52in^2
Or 201in^2 to the nearest whole number.
Rearrange the equation as
cos(x) - √(1 - 3 cos²x) = 0
cos(x) = √(1 - 3 cos²x)
Note that both sides have to be positive, since the square root function is non-negative. If we end up with any solutions that make cos(x) < 0, we must throw them out.
Take the square of both sides.
(cos(x))² = (√(1 - 3 cos²(x))²
cos²(x) = 1 - 3 cos²(x)
4 cos²(x) = 1
cos²(x) = 1/4
cos(x) = ± √(1/4)
cos(x) = ± 1/2
We throw out the negative case and we're left with
cos(x) = 1/2
This has general solution
x = arccos(1/2) + 360° n or x = -arccos(1/2) + 360° n
(where n is any integer)
x = 60° + 360° n or x = -60° + 360° n
Now just pick out the solutions that fall in the interval 0° ≤ x < 360°.
• From the first solution set, we have x = 60° when n = 0.
• From the second set, we have x = 300° when n = 1.
So the answer is D.
