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
Again this is a simple addition problem I think.
Add 3.6+4.705+5.92=
14.225 kilometers that Andrea ran over the three days.
Hope this helps, have a good day. c;
I’m not gonna lie. I’m just answering so I could upload my question
Answer:
The measure of two angles are shown in the diagram. 4x - 11 2x + 5 Which equation can be used to find the value of x? A. 6x - 6 = 180 X. 6 x + 64 = 180 B. 6x - 6 = 360 D. 6x + 84 = 180
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1/2 * n = 2n + 14
Multiply everything by 2 to get rid of the fraction
n = 4n + 28
n - 4n = 28
-3n = 28
n = -9.333333
n = -28/3
