In each table, x increases by 1. We start with x = 0 and stop with x = 3. So we will focus on the y columns of each table as those are different.
Let's move from left to right along the four tables.
For the first table, we go from y = 1 to y = 2. That's an increase of 1
Sticking with the first table, we go from y = 2 to y = 4. The increase is now 2
Since the increase is not the same, this means the table is not linear. The y increase must be constant. We can rule out choice A
Choice B can be ruled out as well. Why? Because...
the jump from y = 0 to y = 1 is +1
the jump from y = 1 to y = 3 is +2
The same problem comes up as it did with choice A
Choice C has the same problem, but the increase turns into a decrease half the time. We go from y = 0 to y = 1, then we go back to y = 0 so the "increase" is really a decrease. We can think of it as a negative increase. Regardless, this allows us to rule out choice C
Only choice D is the answer. Each time x goes up by 1, y goes up by 2. Therefore the slope is 2/1 = 2
Answer:
Step-by-step explanation:
If the dolphins are fed half a bucket of fish then half of that would be a quarter, therefore the sea otters are fed 1/4 of a bucket.
They need to sell 33, unless you round it up then 34.
Answer:
It should take about 4.75 min.
Step-by-step explanation:
Hope this helps and could you mark me as the brainliest
Assuming the order required is as n-> inf.
As n->inf, o(log(n+1)) -> o(log(n)) since the 1 is insignificant compared with n.
We can similarly drop the "1" as n-> inf, the expression becomes log(n^2+1) ->
log(n^2)=2log(n) which is still o(log(n)).
So yes, both are o(log(n)).
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