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KatRina [158]
3 years ago
10

the product of two consecutive integers is 5402. Build a function that can be used to solve for the integers. whatthose two inte

gers.
Mathematics
1 answer:
nasty-shy [4]3 years ago
3 0
Using algebra...
n and n+2 are the integers
n*(n+2)=288
n^2+2n=288
n^2+2n-288=0
288=2*2*2*2*2*3*3=(2*2*2*2)*(2*3*3)=16*18 and 18-16=2
factor
(n+18)(n-16)=0
n+18=0
n=-18
n+2=-16
this is one solution
n-16=0
n=16
n+2=18

this is another solution
as you can see, it's just a matter of factoring
288

using arithmetic...
√288=16.97≈17

16 and 18 are the integers (as well as -16 and -18)
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Hiiiiiiii, okay here's the full- explanation for you, so here's the final answer. The explanation of how i got the answer is also below! Hope this helped you!!

Answer:

2.9

Step-by-step explanation:

First the mean or average = sum / data points (or N) = (2 + 4 + 7 + 8 + 9) / 5 = 6  

Now we take each of the individual values subtract the mean, and then square them, and add it all up.  

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34 / (N - 1) -----> = 34 / (5 - 1) = 8.5  

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4 years ago
What are the domain and range of the function f(x)= X-7+9?
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Answer:

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8 0
3 years ago
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What is the product? x^2-16/2x+8 multiplied by x^3-2x^2+x/x^2+3x-4
Contact [7]

Answer:

(x (x - 4) (x - 1))/(2 (x + 4))

Step-by-step explanation:

Simplify the following:

((x^2 - 16) (x^3 - 2 x^2 + x))/((2 x + 8) (x^2 + 3 x - 4))

The factors of -4 that sum to 3 are 4 and -1. So, x^2 + 3 x - 4 = (x + 4) (x - 1):

((x^2 - 16) (x^3 - 2 x^2 + x))/((x + 4) (x - 1) (2 x + 8))

Factor 2 out of 2 x + 8:

((x^2 - 16) (x^3 - 2 x^2 + x))/(2 (x + 4) (x + 4) (x - 1))

x^2 - 16 = x^2 - 4^2:

((x^2 - 4^2) (x^3 - 2 x^2 + x))/(2 (x + 4) (x + 4) (x - 1))

Factor the difference of two squares. x^2 - 4^2 = (x - 4) (x + 4):

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Factor x out of x^3 - 2 x^2 + x:

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(x - 1) (x - 1) = (x - 1)^2:

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