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Lilit [14]
3 years ago
15

find three consecutive negative integers such that six times the largest is equal to the twice sum of the two smaller integers.

Mathematics
1 answer:
Oksana_A [137]3 years ago
3 0

Answer:

The consecutive negative integers are -3, -4 and -5.

Step-by-step explanation:

Let the three consecutive negative integers be k, (k-1) and (k -2)

Now, according to the question:

6(Largest Integer)  = 2( Sum of smaller two Integers)

or, 6 (k)  = 2( (k-1) + (k-2))

⇒ 6k = 2(2k -3)

or, 6k = 4k - 6

or, 6k - 4k = -6

or, 2k = -6 ⇒  k =  -3

Hence, the largest negative integer = -3

The next two consecutive negative  integer are (k-1) and (k-2)  = (-3-1) and (-3-2)

 = -4 and -5

Hence, the consecutive negative integers are -3, -4 and -5.

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Answer:

x1, x2 = 4.74 , -2.74

Step-by-step explanation:

To find the roots of a quadratic function we have to use the bhaskara formula

ax^2 + bx + c

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a = 1     b = -2    c = -13

x1 = (-b + √ b^2 - 4ac)/2a

x2 =(-b - √ b^2 - 4ac)/2a

x1 = (2 + √ (2^2 - 4 * 1 * (-13)))/2 * 1

x1 = (2 + √ (4 + 52)) / 2

x1 = (2 + √ 56 ) / 2

x1 = (2 + 7.48) / 2

x1 = 9.48 / 2

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x2 = (2 - √ 56 ) / 2

x2 = (2 - 7.48) / 2

x2 = -5.48 / 2

x2 = -2.74

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3 years ago
Two column proofs Beginner's
Darina [25.2K]

LETS prove the 2nd part of the Question

Here concept of right angles is used

We can see BA is perp to BD

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Similarly BC is perp to BE

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as both equations add up gives 90 degree so will equate them

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which means a1 = a3

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To learn more about Right Angles:

brainly.com/question/7116550

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