What are the solution of the equation 2|×+7|-3=5
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Let
x---------> first positive integer
x+1------> second positive integer
x+2-----> third positive integer
we know that
(x+1)*(x+2)=72-------> x² +2x+x+2=72 -------> x² +3x-70=0
using a graph tool-------> <span>I solve the quadratic equation
</span>see the attached figure
the roots are
x1=-10
x2=7
the answer is
first positive integer is x=7
second positive integer is x+1=8
third positive integer is x+2=9
x---------> first positive integer
x+1------> second positive integer
x+2-----> third positive integer
we know that
(x+1)*(x+2)=72-------> x² +2x+x+2=72 -------> x² +3x-70=0
using a graph tool-------> <span>I solve the quadratic equation
</span>see the attached figure
the roots are
x1=-10
x2=7
the answer is
first positive integer is x=7
second positive integer is x+1=8
third positive integer is x+2=9
Candidates range from 1 to 50.
50/4=12 positive integers are multiples of 4
50/6=8 positive integers are multiples of 6
50/12=4 positive integers are multiples of 12 (LCM of 4 and 6)
By the inclusion/exclusion principle, the number of multiples of either 4 or 6 is equal to 12+8-4=16.
Therefore, the complement is the number of positive integers that are multiples of neither 4 nor 6 = 50-16=34.
50/4=12 positive integers are multiples of 4
50/6=8 positive integers are multiples of 6
50/12=4 positive integers are multiples of 12 (LCM of 4 and 6)
By the inclusion/exclusion principle, the number of multiples of either 4 or 6 is equal to 12+8-4=16.
Therefore, the complement is the number of positive integers that are multiples of neither 4 nor 6 = 50-16=34.