How do you reflect over y axis

2 answers:

5 0

Think of a point not on the y-axis.
For example, think of the point (5, 5).
This point is in quadrant 1, right of the y-axis and above the x-axis.
To reflect it over the y-axis, the reflection will be in quadrant 2, still above the x-axis, but now left of the y-axis. The distance the reflection is from the y-axis is the same as the original point is from the y-axis.

The reflection of (5, 5) over the y-axis is (-5, 5).

To reflect any point over the y-axis, replace the x-coordinate with the additive inverse of the x-coordinate.

aleksley [76]3 years ago

3 0

If you want to reflect over the y axis then your answer would want to be going either from the left side to the right side, or right side over to the left side. If the figure was on the top or bottom you would go over the x axis.

So to reflect over the y axis you must find the points that each angle or corner of the figure your reflecting. You'll want to remember those locations of the points, then you'll transfer them over to the other side on the same points but on the other side. I'll make an example to make it more clear

For the first number, before the coma, if the number is negative you'll go left, if the number is positive you go right
For the second number, after the coma, you go up if it's positive or down if its negative.

If A is located at (-3, +2)
Angle B at (-1, +2)
Angle C at (-2, +0)

To transfer them over the y axis the numbers being positive or negative will change
For example

Angle A will become (+3, +2)
Angle B will become (+1, +2)
And angle C will become (+2, 0)

If the number is zero then it stays on the default line.

Hope this helped,reply if you need more help!!

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mezya [45]

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7 0

3 years ago

A right square pyramid with base edges of length $8\sqrt{2}$ units each and slant edges of length 10 units each is cut by a plan

seropon [69]

Answer:

32 $unit^{3}$

Step-by-step explanation:

Given:

The slant length 10 units
A right square pyramid with base edges of length 8 $\sqrt{2}$

Now we use Pythagoras to get the slant height in the middle of each triangle:

$\sqrt{10^{2 - (4\sqrt{2} ^{2} }) }$ = $\sqrt{100 - 32}$ = $\sqrt{68}$ units

One again, you can use Pythagoras again to get the perpendicular height of the entire pyramid.

$\sqrt{68-(4\sqrt{2} ^{2} )}$ = $\sqrt{68 - 32}$ = 6 units.

Because slant edges of length 10 units each is cut by a plane that is parallel to its base and 3 units above its base. So we have the other dementions of the small right square pyramid: