The length of each side of an equilateral triangle is increased by 6 inches, so the perimeter is now 60 inches. What was the ori
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First, rewrite the equation in standard form.
The center-radius form of the circle equation<span> is in the format:
(x – h)^</span>2<span> + (y – k)^</span>2<span> = r^</span>2
<span>with the center being at the </span>point<span> (h, k) and the radius being "r".
</span>
(x-3)^2 + (y+4)^2 = 81
From here, you can determine the center and radius. The center is at (3,-4) and the radius is 9.
The center-radius form of the circle equation<span> is in the format:
(x – h)^</span>2<span> + (y – k)^</span>2<span> = r^</span>2
<span>with the center being at the </span>point<span> (h, k) and the radius being "r".
</span>
(x-3)^2 + (y+4)^2 = 81
From here, you can determine the center and radius. The center is at (3,-4) and the radius is 9.
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.