You’d have to use the Pythagorean Thereom. a^2 + b^2 = c^2. then figure out the smallest side length from there
For this case we have the following original function:
We apply the following transformation:
Vertical displacements:
Assume k> 0
To graph f (x) - k, move the graph k units down.
Applying the transformation for k = 5 we have:
Therefore, the graph moves 5 units down.
Answer:
Third graph
area is 36...............
For three fair six-sided dice, the possible sum of the faces rolled can be any digit from 3 to 18.
For instance the minimum sum occurs when all three dices shows 1 (i.e. 1 + 1 + 1 = 3) and the maximum sum occurs when all three dces shows 6 (i.e. 6 + 6 + 6 = 18).
Thus, there are 16 possible sums when three six-sided dice are rolled.
Therefore, from the pigeonhole principle, <span>the minimum number of times you must throw three fair six-sided dice to ensure that the same sum is rolled twice is 16 + 1 = 17 times.
The pigeonhole principle states that </span><span>if n items are put into m containers, with n > m > 0, then at least one container must contain more than one item.
That is for our case, given that there are 16 possible sums when three six-sided dice is rolled, for there to be two same sums, the number of sums will be greater than 16 and the minimum number greater than 16 is 17.
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Answer:
I would have to say C.14
Step-by-step explanation:
Because if you are running the same exact emperiment with the same exact materials everytime then you should get the same outcome.
Hope this helps! : D
