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Lola A.
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Pre-Algebra
TutorMe
Question:

What are all the possible factors of 80?

Lola A.
Answer:

Remember a factor is a whole number that can divide evenly into another number. Let's start with a simple number like 6. We start with the number 1 and see what other numbers up to 6 can be divided into 6. Note, we can't have something larger than 6 dividing evenly into it. So we stop at 6. The spectrum of numbers we are working with is from 1 ------------> 6 _____________________________________________________________________ FACTORS LIST 6 / 1 = 6 OR 6 = 1 * 6 1, 6 6 / 2 = 3 OR 6 = 2 * 3 2, 3, 6 / 3 = 2 OR 6 = 3 * 2 2, 3, (duplicates, only count once) X 6 / 4 X Does not divide through evenly, so the number 4 is not a factor of 6 X 6 / 5 X Does not divide through evenly, so the number 5 is not a factor of 6 6 / 6 = 1 OR 6 = 6 * 1 The factors of 6 are : 1, 2, 3 and 6 because all these numbers can be divided evenly into 6. The list above is referred to as a factor list. ______________________________________________________________________ So let's solve the question at hand. What are all possible factors of 80: Let's start by creating a factors list. We know the starting point is a 1 and ending point is 80 since we can't have any number larger than 80 dividing into it evenly. So we need to fill in the gap between the number 1 and the number 80 which represent the left and right extremes of the spectrum. SPECTRUM ---------------> 1 to 80 _____________________________________________________________________ FACTORS LIST 80 / 1 = 80 OR 80 = 1 * 80 ------------------> 1, 80 80 / 2 = 40 OR 80 = 2 * 40 -------------------> 2, 40, X 80 / 3 X (3 does not divide into 80 evenly) 80 / 4 = 20 OR 80 = 4 * 20 -------------------> 4, 20, 80 / 5 = 16 OR 80 = 5 * 16---------------------> 5, 16, X 80 / 6 X (6 does not divide into 80 evenly) X 80 / 7 X (7 does not divide into 80 evenly) 80 /8 = 10 OR 80 = 8 * 10-----------------------> 8, 10, X 80 / 9 X (9 does not divide into 80 evenly) 80 / 10 = 8 OR 80 = 10 * 8-----------------------> 8, 10, (only count these factors once) ______________________________________________________________________ From the factors list above I come up with the following numbers: I suggest you travel down the "V" on the left side and back up it on the right side to obtain all the possible factors. You will notice at the bottom of the "V' are duplicates (8 , 10). Only record one set of the duplicate numbers. FINAL ANSWER -------------------------------------------------------> 1, 2, 4, 5. 8, 10, 16, 20, 40 and 80

Basic Math
TutorMe
Question:

The subject under discussion is the Commutative law of addition which states that order does not matter when adding a bunch of things. For example, 5 + 4 + 15 = 9 + 15 = 24 It can also be computed as follows: 5 + 15 + 4 = 20 + 4 = 24 OR 4 + 15 + 5 = 19 + 5 = 24 How many ways can you state the expression, 6 + 8 + 6? And what does it add up to each time?

Lola A.
Answer:

6 + 8 + 6 = 14 + 6 = 20 It can also be expressed as: 6 + 6 + 8 = 12 + 8 = 20 OR expressed as 8 + 6 + 6 = 14 + 6 = 20 So no matter what the order the answer is always the same.

Algebra
TutorMe
Question:

An equation is a statement that two expressions are equal. For example the expression 6+3 equals the expression 7+2 because they are both equal to 9. 6 + 3 = 7 + 2 Whenever we have an equation with a variable in in, we call it an algebraic equation. For example, x +5 = 6 Solve the equation by finding the value of the variable x

Lola A.
Answer:

x + 5 = 6 Look at this and all equations as a scale with two equal sides. Whatever we have on the left side must equal what we have on the right side. If we do something to one side and not the other then we create an inbalance. x + 5 - 5 = 6 - 5 Subtract 5 from the left side Subtract 5 from the right side x + 0 = 1 x = 1 So let's test our hypothesis and see if it is true by going back to the original equation. x + 5 = 6 Now we know that x = 1, we can replace the x with a "1" in the original equation and get the following result. 1 + 5 = 6 Therefore 6 = 6 The problem is now complete.

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