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garik1379 [7]
3 years ago
5

The graph for the equation y=-x+2 is shown. If another equation is graphed so that the system has an infinite number of solution

s, which equation could that be?
A. y=-2(x-1)
B. y=-(x+2)
C. y=-1/4(4x-8)
D. y=-1/2(x+4)
Mathematics
2 answers:
maks197457 [2]3 years ago
5 0

Answer:

C

Step-by-step explanation:

Note that C is exactly -1/4(4x-8) = -1/4*4x -1/4*(-8) = -x+2, so the system with C is the same equation twice, and obviously has an infinite number of solutions.

SOVA2 [1]3 years ago
4 0

Answer: C

Step-by-step explanation:

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The sum of two number is 13 and their product is 42 then what are the two numbers​
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6 and 7

Step-by-step explanation:

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I just listed the two equations out and the answer just popped out. The trick to this is just thinking about two numbers that obviously both below 13 that add up to it.  6 and 7 is the solution.

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3 years ago
Students are doing an experiment: that two fair dice are tossed, and all the outcomes are recorded.Then,an event is described as
Marysya12 [62]

Answer:

1.The random variable X can take any value from( 2,- ----,12) in the event of the sample space.

2.The sample space S = [(1, 1), (1, 2),(2, 1), (1, 3),  (2, 2),(3, 1),  (1, 4),(2, 3),(3, 2),(4, 1), (1, 5),(2, 4), (3, 3),(4, 2),(5, 1),(1, 6),  (2, 5),(3, 4),(4, 3),(5, 2), (6, 1),   (2, 6),(3, 5),   (4, 4), (5, 3), (6, 2),(3, 6), (4, 5),  (5, 4), (6, 3),(4, 6) (5, 5), (6, 4),(5,6),  (6,5),(6,6)]

3. The probability of getting a 4 =   1/12

4. The probability of getting a 12 = 1/36

Step-by-step explanation:

There are 36 outcomes in all and the event shows the possible numbers obtained after the sum .

The event will be E= [ 2,3,4,5,6,7,8,9,10,11,12]

The random variable X can take any value from( 2,- ----,12) in the event of the sample space.

The sample space S = [(1, 1), (1, 2),(2, 1), (1, 3),  (2, 2),(3, 1),  (1, 4),(2, 3),(3, 2),(4, 1), (1, 5),(2, 4), (3, 3),(4, 2),(5, 1),(1, 6),  (2, 5),(3, 4),(4, 3),(5, 2), (6, 1),   (2, 6),(3, 5),   (4, 4), (5, 3), (6, 2),(3, 6), (4, 5),  (5, 4), (6, 3),(4, 6) (5, 5), (6, 4),(5,6),  (6,5),(6,6)]

that is it has all the possible outcomes of both the dice.

The probability of getting a 4= Number of fav outcome/ no of possible outcomes=  3/ 36= 1/12

The probability of getting a 12= Number of fav outcome/ no of possible outcomes=1/36

6 0
2 years ago
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