Let R[x] be the set of polynomials with real coefficients. Define D: R[x] -> R[x] where D is the derivative operator, that is D sends the polynomial, as a function, to its derivative. Which of the following is true? A. D is 1-1 only B. D is onto only C. D is a bijection
If we take any polynomial from R[x], we can always find an antiderivative. That antiderivative, of course, is in R[x] as well. This makes D onto. However, D is not one-to-one because all the constant polynomials will be sent to the zero polynomial under D. Hence, the answer is B.
Ash and his Pikachu are set to battle the next gym leader. However, the gym leader has a unique battle rule. The battle shall consist of five rounds. In each round, each trainer shall send out one Pokemon, and the round commences until a Pokemon faint. When the round ends, each trainer will need to withdraw his Pokemon and send out a new Pokemon. Each round, therefore, will be independent of the other rounds. The gym also has a unique way of awarding gym badges by giving Ash some badges equal to his net score at the end of all five rounds. Ash's net score will be equal to the number of rounds he will win minus the number of rounds he will lose. Ash will not receive any badge if his net score is zero or negative. Assuming that around will never result in a tie and that the probability that Ash will win around is fixed at 2/3. What is the probability that Ash will receive EXACTLY three badges?
Ash will win three badges if he wins four rounds and loses 1 round. With each round having a constant probability of success, and all rounds independent of one another, the problem, therefore, can be modeled by a binomial distribution with n=5 and p=2/3. P(X=4) = 5C4 (2/3)^4 (1/3) = 80/243 Or without referring to the Binomial distribution, we can think of it this way: We want Ash to win four rounds out of the 5. It doesn’t matter which rounds he will win as long as he wins 4. We, therefore, choose 4 of the rounds which Ash can win. This can be done in 5C4 = 5 ways. We are now left with the probability that Ash will indeed win 4 of the rounds. Well, he needs to win four but lose 1. Since the probability of him winning is 2/3 and losing is 1/3, this probability is given by (2/3)^4 *(1/3)^1 = 16/243. Thus, the desired probability is 5(16/243)= = 80/243.
Solve: 2(3x+4)^2 -7(3x+4) = 9.
Just let 3x+4 = y first so that the equation becomes 2y^2 - 7y =9 Now this looks familiar, right? it is a quadratic polynomial equation Now if we "transpose" 9 to the left-hand side we have: 2y^2 -7y -9 =0 We can factor this as (2y-9)(y+1) = 0 and get 2y=9 or y =-1 These yield possible values of y to be y=9/2 and y = -1 Now since 3x+4=y we just plug in our computed y's On one hand, if y =9/2, then 3x+4=9/2 => 3x = 1/2 => x =1/6. On the other hand, if y = -1, 3x+4 = -1 => 3x = -5 => x = -5/3. so the solutions for x are 1/6 and -5/3.