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pentagon [3]
3 years ago
10

Find three consecutive integers have a sum of -36 what is the largest integer

Mathematics
1 answer:
Inessa [10]3 years ago
7 0
3 consecutive integers : x, x + 1, x + 2

x + (x + 1) + (x + 2) = - 36...combine like terms
3x + 3 = -36...subtract 3 from both sides
3x = -36 - 3
3x = - 39....divide both sides by 3
x = -39/3
x = - 13

x + 1 = -13 + 1 = -12
x + 2 = -13 + 2 = -11

the largest integer would be -11
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Answer:

We have to prove

sin⁡(α+β)-sin⁡(α-β)=2 cos⁡ α sin ⁡β

We will take the left hand side to prove it equal to right hand side

So,

=sin⁡(α+β)-sin⁡(α-β)      Eqn 1

We will use the following identities:

sin⁡(α+β)=sin⁡ α cos⁡ β+cos⁡ α sin⁡ β

and

sin⁡(α-β)=sin⁡ α cos ⁡β-cos ⁡α sin ⁡β

Putting the identities in eqn 1

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=[ sin⁡ α cos ⁡β+cos⁡ α sin⁡ β ]-[sin⁡ α cos ⁡β-cos ⁡α sin ⁡β ]

=sin⁡ α cos⁡ β+cos⁡ α sin ⁡β- sin⁡α cos⁡ β+cos ⁡α sin ⁡β

sin⁡α cos⁡β will be cancelled.

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Hence,

sin⁡(α+β)-sin⁡(α-β)=2 cos ⁡α sin ⁡β

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Answer:

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Step-by-step explanation:

Given the geometric sequence

7, 14, 28, ...

We know that a geometric sequence has a constant ratio 'r' and is defined by

a_n=a_1\cdot r^{n-1}

where a₁ is the first term and r is the common ratio

Computing the ratios of all the adjacent terms

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now substituting r = 2 and a₁ = 7 in the nth term

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Therefore, the nth term of the geometric sequence 7, 14, 28, ... is:

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