Answer:
Enjoy and it would be good if you gave more points
Step-by-step explanation:
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
The sum of the angles x and y are 127 degrees.
X + Y = 127
The measure of x is 34 more than half the y.
X = (y/2) + 34
X = (y + 68) / 2
X = 127 - Y
(Y + 68)/2 = 127-Y
254-2Y = Y+68
186 = 3Y
Y = 62
X = 127-62 = 65
The measure of two angles are 62 and 65
Answer:
false
Step-by-step explanation:
rational and set of intergers is out of place
Step-by-step explanation:
5y - 1 = 4y + 5
5y -4y = 5 + 1
Y (1y) = 6
