X^3(1/y^2)-4 is (x^3y^-2)-4 simplified with only positive exponents
Answer: M'(2, - 5), L'(-2, -5), j'(-4, - 1)
Step-by-step explanation:
When we do a reflection over a given line, the distance between all the points (measured perpendicularly to the line) does not change.
The line is y = 1.
Notice that a reflection over a line y = a (for any real value a) only changes the value of the variable y.
Let's reflect the points:
J(-4, 3)
The distance between 3 and 1 is:
D = 3 - 1 = 2.
Then the new value of y must also be at a distance 2 of the line y = 1
1 - 2 = 1
The new point is:
j'(-4, - 1)
L(-2, 7)
The distance between 7 and 1 is:
7 - 1 = 6.
The new value of y will be:
1 - 6 = -5
The new point is:
L'(-2, -5)
M(2,7)
Same as above, the new point will be:
M'(2, - 5)
Answer:
2.8%
Step-by-step explanation:
(7.4 - 7.2) / 7.2 =
0.2 / 7.2 =
2.7777777777778%
