a.) The orientation of ABCD is clockwise, as is the orientation of A'B'C'D'. This means the transformation involves a even number of reflections (may be 0). The orientation of AB is North, and the orientation of A'B' is West, so a rotation of 90° CCW (or equivalent) is involved. We can find the point of intersection of the perpendicular bisectors of AA' and BB' (at (-1, -1)) to determine a suitable center of rotation.
ABCD can be transformed to A'B'C'D' by ...
- rotation 90° CCW about the point (-1, -1)
b.) Rotation by 90° can also be accomplished by reflection across a diagonal line. Since we want the orientation to remain unchanged, we need another reflection to put the figure into its final position. A suitable alternate sequence for mapping ABCD to A'B'C'D' is ...
- reflection across the line y=x
- reflection across the line x=-1
1.5(-10) +29 =
1.5 * (- 10) = -15
-15 + 29 = 14
the answer is 14
Answer:
The work is in the explanation.
Step-by-step explanation:
The sine addition identity is:
.
The sine difference identity is:
.
The cosine addition identity is:
.
The cosine difference identity is:
.
We need to find a way to put some or all of these together to get:
.
So I do notice on the right hand side the 
and the 
.
Let's start there then.
There is a plus sign in between them so let's add those together:
![=[\sin(a+b)]+[\sin(a-b)]](https://tex.z-dn.net/?f=%3D%5B%5Csin%28a%2Bb%29%5D%2B%5B%5Csin%28a-b%29%5D)
![=[\sin(a)\cos(b)+\cos(a)\sin(b)]+[\sin(a)\cos(b)-\cos(a)\sin(b)]](https://tex.z-dn.net/?f=%3D%5B%5Csin%28a%29%5Ccos%28b%29%2B%5Ccos%28a%29%5Csin%28b%29%5D%2B%5B%5Csin%28a%29%5Ccos%28b%29-%5Ccos%28a%29%5Csin%28b%29%5D)
There are two pairs of like terms. I will gather them together so you can see it more clearly:
![=[\sin(a)\cos(b)+\sin(a)\cos(b)]+[\cos(a)\sin(b)-\cos(a)\sin(b)]](https://tex.z-dn.net/?f=%3D%5B%5Csin%28a%29%5Ccos%28b%29%2B%5Csin%28a%29%5Ccos%28b%29%5D%2B%5B%5Ccos%28a%29%5Csin%28b%29-%5Ccos%28a%29%5Csin%28b%29%5D)
So this implies:
Divide both sides by 2:
By the symmetric property we can write:
I don’t know how to sorry
Answer:
Area = 33 inches squared
Step-by-step explanation:
The formula for the area of a kite is:
Area = ½ × (d)1 × (d)2
To find the first diagonal (d1):
d1 = 7 in + 4 in = 11 in
To find the second diagonal (d2):
The triangles on the right of the kite must be 3 4 5 triangle since it is a right triangle with a hypotenuse of 5 and a side of 4, we know half of a diagonal would be 3 inches.
d2 = 3 in + 3 in = 6 in
Now we can plug the diagonals into the area formula.
Area = ½ × 11 × 6 = 33 inches squared
