Answer:
Continuous: g(x) and j(x)
Removable: h(x) and m(x)
Infinite: f(x) and i(x) and k(x)
Jump: l(x)
Step-by-step explanation:
If the unicorn seat on a merry–go–round is 5 feet from the center, and the merry–go–round completes a full rotation in 6 seconds
2 answers:
The answer is 5.23 ft/second
The way I got this answer was first finding the circumference of the merry-go-round using the equation
, where r represents the radius, which is the distance from the center of a circle.
So your equation should look like this
. I used the rounded term of 3.14 for 
Once you multiply those terms, you should get the answer, 31.4. Now this is the total distance around the merry-go-round in feet. So, if it takes 6 seconds to make a full rotation and we are trying to find the amount offeet traveled in one second, we should divide the circumference (31.4) by 6 to give us our answer.
So, 31.4÷6=5.2333...ft/second
And I rounded the answer to the nearest hundredth to get 5.23 feet/second
The way I got this answer was first finding the circumference of the merry-go-round using the equation
So your equation should look like this
Once you multiply those terms, you should get the answer, 31.4. Now this is the total distance around the merry-go-round in feet. So, if it takes 6 seconds to make a full rotation and we are trying to find the amount offeet traveled in one second, we should divide the circumference (31.4) by 6 to give us our answer.
So, 31.4÷6=5.2333...ft/second
And I rounded the answer to the nearest hundredth to get 5.23 feet/second
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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.