I disagree with her because for every 1 is six, it shouldn’t have +y on it that is just adding a number that is unneeded
Answer: 55.041
Step-by-step explanation:
You would take out a 4x out of both terms because this is a common factor
4x^3-36x
4x(x^2-9)
and then you would notice that (x^2-9) is a difference of squares, so that factors out too
4x(x-3)(x+3)
that is all you can factor! hope this helps
Let x be the 1st odd number, and x+2 the second odd consecutive number:
(x)(x + 2) = 6[((x) + (x+2)] -1
x² + 2x = 6(2x + 2) - 1
x² + 2x = 12x +12 - 1
And x² - 10x - 11=0
Solve this quadratic expression:
x' = [+10 +√(10²- 4.(1)(-11)]/2 and x" = [+10 -√(10²- 4.(1)(-11)]/2
x' = [10 + √144]/2 and x" = [10 - √64]/2
x' = (10+12)/2 and x" = (10-12)/2
x = 11 and x = -1
We have 2 solutions that satisfy the problem:
1st for x = 11, the numbers at 11 and 13
2nd for x = - 1 , the numbers are -1 and +1
If you plug each one in the original equation :(x)(x + 2) = 6[((x) + (x+2)] -1
you will find that both generates an equlity
Answer:
the answer will be 10 aka number 1
Step-by-step explanation:
If the left side is 18 and the number above it is half that aka 9 then the other side is 10
