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djverab [1.8K]
3 years ago
5

What is y=x^4/x^3 when values of x= {-2, 0, 3, 5, 7}

Mathematics
1 answer:
VMariaS [17]3 years ago
3 0

For this case we have a function of the formy = f (x) where:

f (x) = \frac {x ^ 4} {x ^ 3}

We simplify eliminating common terms of the numerator and denominator:

f (x) = x

We must find the values of the function when:

x = {-2, 0, 3, 5, 7}

So, we have:

For x = -2:

f (-2) = - 2

For x = 0

f (0) = 0

For x = 3

f (3) = 3

For x = 5

f (5) = 5

For x = 7:

f (7) = 7

Thus, the values of the function are:

y = {-2, 0, 3, 5, 7}

Answer:

y = {-2, 0, 3, 5, 7}

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4 0
1 year ago
ANYONE, PLEASE ANSWER ALL THESE QUESTIONS AND I WILL MARK HIM/HER BRAINLIEST + 57 PTS.
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Answer:

1.B

2.C

3.D

4.

a)1/9

b)5/12

5.2/5

Step-by-step explanation:

1.

Since there are 15 total outcomes of which 6 are desirable, the probability of randomly getting a red ball is 5/16 or choice B.

2.

Since there are a total of 8+7+6=21 outcomes, out of which 7 are neither red nor green, there is a 7/21=1/3 probability of that outcome, or answer choice C.

3.

Out of the 52 possible outcomes, 12 are face cards. Therefore, the probability of drawing a face card is 12/52=3/13, or answer choice D.

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For each die, there are a total of 6 possible outcomes. This means that in total there are 6*6=36 possible combinations. Now, let's do some casework.

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b)

All of the possible prime numbers that can be formed with the dice are 2, 3, 5, 7, and 11. For 2, there is only one possible situation where this can happen. For 3, there are 2 possible. For 5, there is 4-1, 3-2, 2-3, and 1-4, or 4 possibilities. For 7, there is 1-6, 2-5, 3-4, 4-3, 5-2, and 6-1, or 6 total possibilities. Finally, for 11, there is only 5-6 and 6-5, or 2 possible. In total, the probability of getting a prime sum is 15/36=5/12.

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