Answer: 15n³-105n²+2n+16/6n²-42n
Step-by-step explanation:
n+8/3n²-21n +5n/2
2(n+8)+5n(3n²-21n)/2(3n²-21n)
2n+16+15n³-105n²/6n²-42n
15n³-105n²+2n+16/6n²-42n
12x^3-11x^2+9x+18 divided by 4x+3
put he division into fraction
12x^3-11x^2+9x+18/4 x +3
reduced fraction by 2
12x^3-11x^2+9x+9/2 x +3
calculate sum
12x^3-11x^2+27/2 x +3
that is your answer ^
hope this helps :)
9514 1404 393
Answer:
q = 40
Step-by-step explanation:
When the quadratic has roots p and r, it can be factored as ...
(x -p)(x -r) = x² -(p+r)x +pr
So, the sum of the roots is 14, and their difference is 6. This lets us find the roots from ...
p + r = 14
p - r = 6
2p = 20 . . . add the two equations
p = 10
r = 14 -p = 4
The value of interest is then ...
q = pr = (10)(4)
q = 40
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The graph shows the roots to be 4 and 10, as we found.
Don't touch the center. It is already even.
Start anywhere by connecting a dotted line from one vertex to the next. To keep things so we know what we are talking about, go clockwise. Now you have 2 points that are Eulerized that were not before.
Skip and edge and do the same thing to the next two vertices. Those two become eulerized. Skip an edge and do the last 2.
Let's try to describe this better. Start at any vertex and number them 1 to 6 clockwise.
Join 1 to 2
Join 3 to 4
Join 5 to 6
I think 3 is the minimum.
3 <<<< answer
