The product of two consecutive positive integers is 1111 more than their sum. find the integers.
1 answer:
The correct question is
<span>The product of two consecutive positive integers is 11 more than their sum. find the integers.
Let
x-------> the first </span>positive integer
x+1-----> the second consecutive positive integer
we know that
x*(x+1)=11+(x+(x+1))----> x²+x=11+(2x+1)----> x²+x=2x+12
x²-x-12=0
using a graph tool------> to resolve the second order equation
see the attached figure
the solution is
x=4
therefore
the numbers are
x=4
x+1=5
the answer is
the numbers are 4 and 5
<span>The product of two consecutive positive integers is 11 more than their sum. find the integers.
Let
x-------> the first </span>positive integer
x+1-----> the second consecutive positive integer
we know that
x*(x+1)=11+(x+(x+1))----> x²+x=11+(2x+1)----> x²+x=2x+12
x²-x-12=0
using a graph tool------> to resolve the second order equation
see the attached figure
the solution is
x=4
therefore
the numbers are
x=4
x+1=5
the answer is
the numbers are 4 and 5
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Answer:
The 38th term of 459,450,441,.. will be:
Step-by-step explanation:
Given the sequence
An arithmetic sequence has a constant difference 'd' and is defined by
computing the differences of all the adjacent terms
so
The first element of the sequence is
so the nth term will be
Putting n=38 to find the 38th term
Therefore, the 38th term of 459,450,441,.. will be: