the assumption being that the first machine is the one on the left-hand-side and the second is the one on the right-hand-side.
the input goes to the 1st machine and the output of that goes to the 2nd machine.
a)
if she uses and input of 6 on the 2nd one, the result will be 6² - 6 = 30, if we feed that to the 1st one the result will be √( 30 - 5) = √25 = 5, so, simply having the machines swap places will work to get a final output of 5.
b)
clearly we can never get an output of -5 from a square root, however we can from the quadratic one, the 2nd machine/equation.
let's check something, we need a -5 on the 2nd, so

so if we use a "1" as the output on the first machine, we should be able to find out what input we need, let's do that.

so if we use an input of 6 on the first machine, we should be able to get a -5 as final output from the 2nd machine.

D. One reason, The others questions don't add up. Reason being, the perimeter of the garden is 4x+60, all of the other questions don't make up for that answer.
(I also can't read D, so I'm just assuming now.)
Answer:
A.
Step-by-step explanation:
1 radian = 180/π
So
Multiplying both sides by 25π/18
We'll get,
25π/18 (r) = (180/π)×(25π/18)
= (180×25)/18
= 10×25
= 250 degrees
Answer:
9:1
Step-by-step explanation:
432/48=9
9:1