Ooooppppppp— homie not gettin nun at home I see
Answer:
The answer in the procedure
Step-by-step explanation:
we know that
The rule of the reflection of a point over the y-axis is equal to
A(x,y) ----->A'(-x,y)
That means -----> The x-coordinate of the image is equal to the x-coordinate of the pre-image multiplied by -1 and the y-coordinate of both points (pre-image and image) is the same
so
A(3,-1) ------> A'(-3,-1)
The distance from A to the y-axis is equal to the distance from A' to the y-axis (is equidistant)
therefore
To reflect a point over the y-axis
Construct a line from A perpendicular to the y-axis, determine the distance from A to the y-axis along this perpendicular line, find a new point on the other side of the y-axis that is equidistant from the y-axis
Fifth grade has 24 students per each teacher.
Sixth grade has about 23 students per teacher.
Sixth grade is lower.
Answer:
2y-2x-2y= - 22x = 0x + 2y = 1
Step-by-step explanation:
1) 462/11= 42
42x7=294
2) 9.38/7= $1.34
3) 6/11= 0.54545454545455
4) 1/4= 0.25
2/9= 0.22222222222
3/10= 0.30
2/1= 2.00
1/5= 0.20
least to greatest- 1/5, 2/9, 1/4, 3/10, 2/1
5) 9/15= 0.6 > 60%
to turn into a percent, move 2 decimal places!
6) 10/4= 2.5
2.5x18= 45
7) 2/3= 0.66666666666
0.66666666666x7= 4.66666666667
