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Nana76 [90]
2 years ago
15

If p(x)=x^2+x+1 and q(x)=3x^2-1, find p(7)

Mathematics
1 answer:
Delvig [45]2 years ago
7 0

Answer:

p(7) = 57

Step-by-step explanation:

Because this is a function, all you have to do is input 7 for x like so:

p(x)=x^2+x+1 --------------->     p(7)=7^2+7+1

                                            so 49 + 8

                                                = 57

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Find two positive consecutive odd integers such that square of the smaller integer is 10 more than the larger integer
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Let 2n+1 be the smaller integer. The larger integer is then 2n+3, and we have

(2n+1)^2=10+(2n+3)\implies4n^2+4n+1=2n+13

\implies4n^2+2n-12=0

\implies2n^2+n-6=0

\implies(2n-3)(n+2)=0

\implies 2n-3=0\text{ or }n+2=0

\implies n=\dfrac32\text{ or }n=-2

We omit n=-2, since 2(-2)+1=-3 is negative.

Then for n=\dfrac32 we find 2\left(\dfrac32\right)+1=4, but this is not odd.

There are no consecutive odd integers that satisfy the given condition!

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3 years ago
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Answer : The fifth term in the sequence is -2.

Step-by-step explanation :

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Answer:

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Read 2 more answers
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