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:
5083
Step-by-step explanation:
6^3 + 5(8 + 15)
Simplify the bracket first
6 x 6 x 6 + 5 (23)
216 + 5 (23)
221 (23)
221 x 23 = 5083
Step-by-step explanation:
1) Ans 3
2) Ans 6
3) Ans 4
4) Ans 1/5
5) Ans 1/23
Answer:
100/8 = 12.5 (s)
Step-by-step explanation:
hmm bro
Answer:
93
Step-by-step explanation:
the total of all those are 150 so if you multiply it by three you get 450
so multiply pizza by three too
to get 93
