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Alisiya [41]
3 years ago
5

Suppose Set B contains 69 elements and the total number elements in either Set A or Set B is 107. If the Sets A and B have 13 el

ements in common, how many elements are contained in set A?
Mathematics
1 answer:
erastovalidia [21]3 years ago
4 0
Set A or B refers to the union of A and B, in other words how many element there are in TOTAL. We are told Set B has 69 elements, but that 13 of them overlap with Set A. This means Set B has 69-13=56 unique elements, and all others are therefore part of set A. If there are 107 elements in total, Set A has 107-56=51 elements.
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vagabundo [1.1K]

Answer:

Considering the given equation y = log_{3}x\\

And the ordered pairs in the format (x, y)

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For (1, a_{1} )

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So the ordered pair will be (1, 0 )

For (3, a_{2} )

x=3

y=a_{2}

y = log_{3}x\\y = log_{3}3\\y = 1

So the ordered pair will be (3, 1 )

For (9, a_{3} )

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y = log_{3}x\\y = log_{3}9\\y=2\log _3\left(3\right)\\y=2

So the ordered pair will be (9, 2 )

For (27, a_{4} )

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So the ordered pair will be (27, 3 )

For (81, a_{5} )

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4 years ago
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Answer:

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input y = kx +6 into the equation of the curve x² + y² – 10x + 8y = 84

x² + (kx + 6)² - 10x + 8(kx + 6) = 84

expand:

x² + k²x² + 12kx + 36 - 10x + 8kx + 48 = 84

simplify by collecting like terms:

x² + k²x² + 20kx - 10x + 84 = 84

subtract 84 on both sides to bring it to the left:

x² + k²x² + 20kx - 10x + 84 - 84 = 0

x² + k²x² + 20kx - 10x = 0

factorise out x:

x²(1 + k²) + x(20k - 10) = 0

using the discriminant b² - 4ac where b is 20k - 10, a is 1 + k² and c is 0, substitute them in the formula b² - 4ac:

b² - 4ac

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the part highlighted in bold is gone because it's all multiplied by 0, so we are left with (20k - 10)² = 0

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20k = 10 and 20k = 10

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k = 1/2

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