X(A')=X(A)+4= -1+4=3
X(B')=X(B)+4= -1+4=3
X(C')=X(C)+4= 3+4=7
4 points to the right
Different parallelograms require different ways of solving so theorems tell you how each parallelogram can be solved
We know that sin2x=2sinxcosx
(search the net for proof if you wish)
So the original equation becomes
2sinxcosx-sinx=0
The two terms both have sinx that can be taken out to get:
sinx(2cosx-1)=0
This is true if sinx=0 or 2cosx-1=0 , rewritten: cosx=1/2
sinx=0 than x=2kπ
cosx=1/2 than x=π/3+2kπ
where k is an integer
The best and most correct answer among the choices provided by your question is the second choice or letter B.
Reflection of a horizontal line <span>has the same result as a rotation of 90 degrees clockwise.</span>
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Yes. Take for example a square and an ellipsis with the same perimeter. The family of ellipses with the same perimeter can have any area between that of a circle to zero (if it is extremely “thin” i.e. if its eccentricity is large). The circle has the maximum area of any other shape with the same perimeter, so the square has the same area of one of the intermediate ellipses.
