I assume you mean pi.
How many different ways can an isosceles trapezoid be reflected across a line of symmetry so that it carries onto itself?
1 answer:
You might be interested in
Rx - sx + y = b
WHEN SOLVING FOR X :
rx - sx + y = b
We must get x onto it's own side, so subtract y from both side.s
rx - sx = b - y
Then, factor out x.
x(r - s) = b - y
Then, divide both sides by (r - s).
x(r - s) ÷ (r - s) = b - y ÷ (r - s)
Simplify.
x = b - y / r - s →
WHEN SOLVING FOR Y :
rx - sx + y = b
We need to isolate y, so get rid of everything BUT y on the left side.
Subtract rx from both sides.
-sx + y = b - rx
Then, add sx to both sides.
y = b - rx + sx
~Hope I helped!~
WHEN SOLVING FOR X :
rx - sx + y = b
We must get x onto it's own side, so subtract y from both side.s
rx - sx = b - y
Then, factor out x.
x(r - s) = b - y
Then, divide both sides by (r - s).
x(r - s) ÷ (r - s) = b - y ÷ (r - s)
Simplify.
x = b - y / r - s →
WHEN SOLVING FOR Y :
rx - sx + y = b
We need to isolate y, so get rid of everything BUT y on the left side.
Subtract rx from both sides.
-sx + y = b - rx
Then, add sx to both sides.
y = b - rx + sx
~Hope I helped!~