The factor of the function will be (x - 1), (x - 2), and (x + 2). Then the roots of the function will be 1, 2, and negative 2.
<h3>What is a factorization?</h3>
It is a method for dividing a polynomial into pieces that will be multiplied together. At this moment, the polynomial's value will be zero.
The function is given below.
f(x) = 4x³ - 4x² - 16x + 16
Then the factor of the function will be
f(x) = 4x²(x - 1) - 16(x - 1)
f(x) = 4(x - 1) (x² - 4)
f(x) = 4(x - 1) (x² - 2²)
f(x) = 4(x - 1)(x - 2)(x + 2)
Then the roots of the function will be
x = 1, 2, -2
More about the factorization link is given below.
brainly.com/question/6810544
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A standard deck has 52 cards.
A standard deck has 4 jacks.
A standard deck has 13 clubs.
From this, we can derive the following:
The probability of drawing a jack is 4/52 or 1/13
The probability of drawing a club is 13/52 or 1/4
But since the problem asks for drawing jack or club, therefore we should add the 2 probabilities, making 17/52. This is not the final answer yet. We know that there is a jack of clubs, therefore we need to subtract 1 from the probabilities since jack of clubs were considered in the 2 categories of probability.
With that being said, the probability of drawing a club or a jack is 16/52 or 4/13
Bonus Question:
The first thing you need to do here is find the probability of each scenario. First let's do what is given, the probability of drawing 2 aces. Since there are 4 aces in a deck of 52, we can easily say that the probability of drawing an ace is 4/52. However for our second draw, the probability of drawing a different ace is 3/51. This is so since we already drew a card that is an ace, hence we need to subtract one from the total aces (4-1) and from the total cards in the deck (52-1). In getting the probability of drawing two aces, we need to multiply the said probabilities: 4/52 and 3/51, resulting to 1/221.
For the second scenario, the drawing of 2 red cards, we just use the same concept but in this, we are already considering the 2 red cards in the first scenario, therefore the chance of drawing a red on our first draw is 24/52. For our second, we just need to subtract one card, therefore 23/51. Multiply these two and we will get 46/221.
Now, the problem asks for the chance of drawing either 2 reds or 2 aces, therefore we add the probabilities of the 2 scenarios:
46/222 + 1/221 = 47/221
Summary:
First Scenario:
4/52 + 3/51 = 1/221
Second Scenario:
24/52 + 23/51 = 46/221
Chances of drawing 2 red cards or 2 aces:
1/221 + 46/221 = 47/221
Answer: A) max at (14, 6) = 64, min at (0,0) = 0
<u>Step-by-step explanation:</u>
Graph the lines at look for the points of intersection.
Input those points into the Constraint function (2x + 6y) and look for the maximum value and minimum value.
Points of Intersection: (0, 0), (17, 0), (0, 10), (14, 6)
Point Constraint 2x + 6y
(0, 0): 2(0) + 6(0) = 0 Minimum
(17, 0): 2(17) + 6(0) = 34
(0, 10): 2(0) + 6(10) = 60
(14, 6): 2(14) + 6(6) = 64 Maximum
Answer:
8
Step-by-step explanation:
Just count the spaces between them.
