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ivann1987 [24]
3 years ago
11

The 6 consecutive integers is 393 what is the third number in this sequence

Mathematics
1 answer:
MAXImum [283]3 years ago
3 0
X+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)=393
6x+15=393
6x=393-15
6x=378
x=63
third one is 63+2=65
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3 years ago
What is the product of all constants $k$ such that the quadratic $x^2 + kx +15$ can be factored in the form $(x+a)(x+b)$, where
Setler79 [48]

Answer:

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Step-by-step explanation:

We are given quadratic equations as

x^2+kx+15

and it can be factored as

=(x+a)(x+b)

now, we can multiply factor term

(x+a)(x+b)=x^2+(a+b)x+ab

now, we can compare

x^2+(a+b)x+ab=x^2+kx+15

so, we get

k=a+b

ab=15

we are given that

'a' and 'b' are integers

so, we can find all possible factors

15=(-1\times -15),(1\times 15)

15=(-3\times -5),(3\times 5)

so, we can find k

At (-1\times -15):

k=a+b

we can plug values

k=-1-15

k=-16

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k=a+b

we can plug values

k=1+15

k=16

At (-3\times -5):

k=a+b

we can plug values

k=-3-5

k=-8

At (3\times 5):

k=a+b

we can plug values

k=3+5

k=8

So, values of k are

k=-16,k=-8,k=8,k=16

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Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Can you help me with this question ​
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Answer:

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Step-by-step explanation:

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3 years ago
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