I'm assuming that the diagonals intersect at E. Since diagonals of a parallelogram bisect each other, BE=ED.
- 7x-2=x²-10
- x²-7x-8=0
- (x-8)(x+1)=0
- x = -1, 8
As distance must be positive, we reject the negative case, so x=8.
Thus, BE=ED=54.
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Let's say that we don't have jokers. In that case, there are 52 cards, and half of them are red (so 26 are red). The probability of pulling a red card once is 1/2 since half the cards are red.
If we pick one out, there are only 51 total cards left and 25 red cards. So, the probability of picking one again would be 25/51.
We multiply the probabilities of these two events to find the probability of them both happening.
1/2×25/51=25/102
The probability of picking two red cards in a row is 25/102 or around 24.5%.
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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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