What set of reflections would carry trapezoid abcd onto itself?

a. y=x, x-axis, y=x, y-axis/p>
b. x-axis, y=x, x-axis, y=x
c. y-axis, x-axis, y-axis
d. x-axis, y-axis, y-xis

a. y=x, x-axis, y=x, y-axis

Step-by-step explanation:

We have,

'Reflection across x-axis changes (x,y) to (x,-y)'

'Reflection across y-axis changes (x,y) to (-x,y)'

'Reflection across y=x changes (x,y) to (y,x)'

As, ABCD have co-ordinates A(1,-1), B(4,-1), C(2,-2) and D(3,-2).

We see that, on applying the reflections in the sequence 'y=x, x-axis, y=x, y-axis', the following table is obtained,

                   y=x             x-axis           y=x             y-axis

A=(1,-1)        (-1,1)             (-1,-1)             (-1,-1)             (1,-1)

B=(4,-1)       (-1,4)            (-1,-4)             (-4,-1)            (4,-1)

C=(2,-2)      (-2,2)           (-2,-2)            (-2,-2)          (2,-2)

D=(3,-2)      (-2,3)            (-2,-3)            (-3,-2)          (3,-2)

Hence, the sequence 'y=x, x-axis, y=x, y-axis' maps ABCD onto itself.

i believe the answer is 68cm

step-by-step explanation:

i just took 34 x 2 because the space between v and y is the same as y to k.

13

step-by-step explanation:

169-1=168

168-3=165

165-5=160

160-7=153

153-9=144

144-11=133

133-13=120

120-15=105

105-17=88

88-19=69

69-21=48

48-23=25

25-25=0

since it took 13times to completely subtract it hence its square root is 13