Multiply both sides by c
X = d*c
by the double angle identity for sine. Move everything to one side and factor out the cosine term.
Now the zero product property tells us that there are two cases where this is true,
In the first equation, cosine becomes zero whenever its argument is an odd integer multiple of
, so
where
![n[/tex ]is any integer.\\Meanwhile,\\[tex]10\sin x-3=0\implies\sin x=\dfrac3{10}](https://tex.z-dn.net/?f=n%5B%2Ftex%20%5Dis%20any%20integer.%5C%5CMeanwhile%2C%5C%5C%5Btex%5D10%5Csin%20x-3%3D0%5Cimplies%5Csin%20x%3D%5Cdfrac3%7B10%7D)
which occurs twice in the interval
for
and
. More generally, if you think of
as a point on the unit circle, this occurs whenever
also completes a full revolution about the origin. This means for any integer
, the general solution in this case would be
and
.
4/6 or 2/3. All you need to do is add the amount of blue and red sides from the cube and divide it by the total number of sides on a cube.
You can also say 66%
Hope this helps!! :)
Minus
The quadratic equation is
( -b+-SQRT(b^2 - 4ac) )/2a
Choice C for problem 6 is correct. The two angles (65 and 25) add to 90 degrees, proving they are complementary angles.
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The answer to problem 7 is also choice C and here's why
To find the midpoint, we add up the x coordinates and divide by 2. The two points A(-5,3) and B(3,3) have x coordinates of -5 and 3 respectively. They add to -5+3 = -2 which cuts in half to get -1. This means C has to be the answer as it's the only choice with x = -1 as an x coordinate.
Let's keep going to find the y coordinate of the midpoint. The points A(-5,3) and B(3,3) have y coordinates of y = 3 and y = 3, they add to 3+3 = 6 which cuts in half to get 3. The midpoint has the same y coordinate as the other two points
So that is why the midpoint is (-1,3)
