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klemol [59]
3 years ago
15

If the point of the preimage is (5,2) and the point of the image is (1,2), what RULE was applied to the preimage to obtain the i

mage?
A) (x, y + 4)
B) (x + 4, y)
C) (x - 4, y)
D) (x, y - 4)
Mathematics
1 answer:
marusya05 [52]3 years ago
5 0
<h3>Answer: Choice C)  (x-4, y)</h3>

Explanation:

We have x = 5 turn into x = 1. We need to subtract 4 from the x coordinate to get this. Therefore x turns into x-4.

The y coordinate stays at y = 2 the entire time, so y isn't altered. We have y map to y.

Overall, the rule is (x,y) \to (x-4, y) which translates the point 4 units to the left.

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Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl; then a seco
kari74 [83]

Answer:

C. \frac{1}{18}

Step-by-step explanation:

Given: Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl then a second card is drawn.

To Find: If the cards are drawn at random and if the sum of the numbers on the cards is 8, what is the probability that one of the two cards drawn is numbered 5.

Solution:

Sample space for sum of cards when two cards are drawn at random is \{(1,1),(1,2),(1,3)......(6,6)\}

total number of possible cases =36

Sample space when sum of cards is 8 is \{(3,5),(5,3),(6,2),(2,6),(4,4)\}

Total number of possible cases =5

Sample space when one of the cards is 5 is \{(5,3),(3,5)\}

Total number of possible cases =2

Let A be the event that sum of cards is 8

p(\text{A}) =\frac{\text{total cases when sum of cards is 8}}{\text{all possible cases}}

p(\text{A})=\frac{5}{36}

Let B be the event when one of the two cards is 5

probability than one of two cards is 5 when sum of cards is 8

p(\frac{\text{B}}{\text{A}})=\frac{\text{total case when one of the number is 5}}{\text{total case when sum is 8}}

p(\frac{\text{B}}{\text{A}})=\frac{2}{5}

Now,

probability that sum of cards 8 is and one of cards is 5

p(\text{A and B}=p(\text{A})\times p(\frac{\text{B}}{\text{A}})

p(\text{A and B})=\frac{5}{36}\times\frac{2}{5}

p(\text{A and B})=\frac{1}{18}

if sum of cards is 8 then probability that one of the cards is 8 is \frac{1}{18}, option C is correct.

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Answer:

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Step-by-step explanation:

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Answer: y(-3y^7+18y^4+10)

Step-by-step explanation:

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4 more than Manuel did
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