The 4 consecutive of integers greater than 9 (but not including 9) are 11, 13, 15, 17. Added together, we get 56.

The 4 consecutive of integers greater than 9 (but not including 9) are 11, 13, 15, 17. Added together, we get 56.

An odd integer can be expressed as because two times any number is an even number and one plus an even number is always odd. We can then write these three consecutive odd integers in terms of as . We can then square each of these numbers and add them together.

Then use binomial expansion to rewrite the expression (better known as FOIL).

We can then combine like terms and set it equal to as given.

This tells us that two possible sets of numbers satisfy this condition: and . It is evident that the sums of the squares of these numbers should be the same, so we cannot determine which set the question is discussing.

An odd integer can be expressed as because two times any number is an even number and one plus an even number is always odd. We can then write these three consecutive odd integers in terms of as . We can then square each of these numbers and add them together.

Then use binomial expansion to rewrite the expression (better known as FOIL).

We can then combine like terms and set it equal to as given.

This tells us that two possible sets of numbers satisfy this condition: and . It is evident that the sums of the squares of these numbers should be the same, so we cannot determine which set the question is discussing.

For a problem like this, you can always use the answers to find your correct answer. By choosing each number, you can find the other two options and then add together your values. You would, for instance, take and say, "The other two must be and ." Then, adding them to get , you will know that this is not correct.

However, you can do this much more easily with algebra. You know that three consecutive integers are going to look like:

, where is the price of the least expensive candy. Thus, you know that the total price of your candies can be represented in the following manner:

This simplifies to:

Solving for , you get:

Remember that you need to find the highest priced candy. Therefore, the answer is or .

For a problem like this, you can always use the answers to find your correct answer. By choosing each number, you can find the other two options and then add together your values. You would, for instance, take and say, "The other two must be and ." Then, adding them to get , you will know that this is not correct.

However, you can do this much more easily with algebra. You know that three consecutive integers are going to look like:

, where is the price of the least expensive candy. Thus, you know that the total price of your candies can be represented in the following manner:

This simplifies to:

Solving for , you get:

Remember that you need to find the highest priced candy. Therefore, the answer is or .

to find the integers you can guess and check (you know both are larger than 10 because their product is greater than 100) or you can set up a system of equations. if a is the larger number and .

Therefore:

if you solve that quadratic you get

and b is the smaller number so the bigger number is 12

to find the integers you can guess and check (you know both are larger than 10 because their product is greater than 100) or you can set up a system of equations. if a is the larger number and .

Therefore:

if you solve that quadratic you get

and b is the smaller number so the bigger number is 12

A geometric sequence is one where two get two each consecutive number in the sequence, you must multiply or divide a number. If we look at the sequence, we can see that the pattern is dividing by each time. Therefore, to get the next term in the sequene, we must divide the last term given in the sequence:

A geometric sequence is one where two get two each consecutive number in the sequence, you must multiply or divide a number. If we look at the sequence, we can see that the pattern is dividing by each time. Therefore, to get the next term in the sequene, we must divide the last term given in the sequence: