To square means to multiply something by itself. A variable is a letter such as x. So a squared variable=x×x=x^2.
I hope that helped.
The first thing you should do is to determine all the zeros, and you will have 7, -11, 2 + 6i, 2 - 6i.
After that you have to <span>subtract and (x) from every each zero that you have above, which means
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<span>And the last step is to multiply them together, and that's all you need!
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<span>I am pretty sure that everything hase become clear! Regards.</span>
If Deliah does jumping jacks at a constant rate, this means that she does them at the same pace or you could say that she does the same amount of jumping jacks in a specified amount of time, ie. if you counted how many jumping jacks she did in one minute, it would be same as how many she would complete in the next minute, and the next, and so on.
Now given that she does 184 jumping jacks in four minutes, and she has kept a constant pace throughout, to find out how many she does each minute, we simply need to divide the number of jumping jacks she does in 4 minutes by 4. Thus:
Jumping jacks in 1 minute = Jumping jacks in 4 minutes / 4
= 184 / 4
= 46
Thus, Deliah can do 46 jumping jacks per minute.
The box and whisker plot is attached.
We first order the data from least to greatest:
6, 7, 11, 13, 14, 15, 15, 19, 21
The median is the middle value, or 14.
The lower quartile is the median of the lower half (split by the median). This is between 7 and 11: (7+11)/2 = 18/2 = 9
The upper quartile is the median of the upper half (split by the median). This is between 15 and 19: (15+19)/2 = 34/2 = 17
The highest value is 21.
The lowest value is 6.
We draw the middle line of the box at 14, the median. We draw the left side of the box at the lower quartile, 9. We draw the right side of the box at the upper quartile, 17. From the right side of the box, we draw a whisker to the highest value, 21. From the left side of the box, we draw a whisker to the lowest value, 6.
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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