Name the set of 6 consecutive integers starting with -3. (Put the set in braces { } and put commas between the elements of the set.)
The product of four consecutive integers is always one less than a perfect square. The product of four consecutive integers starting with n will be one less than the square of n2 + 3n + 1
0 and 1. Integers do not include decimals, where as 9% is equal to 0.09. So the consecutive integers would be 0 and 1.
None. But, 14 is between the two odd integers 13 and 15.
The square of 3 is 9, which does not lie between consecutive integers. Perhaps you mean the square root of 3, which lies between 1 and 2.
12and 11
"Consecutive" integers are integers that have no other integer between them.
{3, 4, 5, 6, 7, 8}. Done!
two consecutive integers of the square root of 66 found between
The product of four consecutive integers is always one less than a perfect square. The product of four consecutive integers starting with n will be one less than the square of n2 + 3n + 1
There are 30 such integers.
between -6 and -7
0 and 1. Integers do not include decimals, where as 9% is equal to 0.09. So the consecutive integers would be 0 and 1.
None. But, 14 is between the two odd integers 13 and 15.
138 is, itself, an integer. It is impossible for any integer to lie between two consecutive integers.
31.how do you solve?
There is no difference, only your outcome. The formula for both is x+2
59 and 61