Given that
z₁ = 15 (cos(90°) + i sin(90°))
z₂ = 3 (cos(10°) + i sin(80°))
we get the quotient z₁/z₂ by dividing the moduli and subtracting the arguments:
z₁/z₂ = 15/3 (cos(90° - 10°) + i sin(90° - 10°))
z₁/z₂ = 5 (cos(80°) + i sin(80°))
so that z₁ is scaled by a factor of 1/3 and is rotated 10° clockwise.
Answer:
ok
Step-by-step explanation:
Answer:
Matrix transformation = ![\left[\begin{array}{ccc}-1&0\\0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%260%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D)
Vertices of the new image: P'= (5,-2), Q'= (6,-3), R'= (2,-3)
Step-by-step explanation:
Transformation by reflection will produce a new congruent object in different coordinate. Reflection to y-axis made by multiplying the x coordinate with -1 and keep the y coordinate unchanged. The matrix transformation for reflection across y-axis should be:
.
To find the coordinate of the vertices after transformation, you have to multiply the vertices with the matrix. The calculation of the each vertice will be:
P'=
= (5,-2)
Q'=
= (6,-3)
R'=
= (2,-3)
Answer:
28-11=17
28=100%
17=x
17×100/28=60,71
rounded 61%
60,71%
11/28*100=a
a-100=60,71
to nearest percent 61 i think
Answer:
1196 mm^2
Step-by-step explanation:
That's really well described. All in all you have 6 squares all laid out flat. Each square has a side of 14 mm.
The area of 1 square = 14mm * 14mm = 196 mm^2
The area of 6 squares = 6 * 196 mm^2 = 1176 mm^2