Sign In
| 13 Theory slides |
| 11 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
Here are a few recommended readings before getting started with this lesson.
Ignacio got a part-time job at a restaurant. His first task is to peel some garlic. Ms. Ley, the restaurant owner, tells him to peel all the heads of garlic in a crate.
Commutative Property of Multiplication
Associative Property of Multiplication
am⋅an=am+n
Commutative Property of Multiplication
am⋅an=am+n
Multiply
Ignacio is helping Ms. Ley make hand-pulled noodles.
To encourage Ignacio to keep doing a great job, Ms. Ley gives him a bonus each time he gets a compliment from a customer.
Associative Property of Multiplication
Commutative Property of Multiplication
a⋅a=a2
Associative Property of Multiplication
am⋅an=am+n
Multiply
Everyone has bad days and today was no exception. In addition to the four compliments from the customers, Ignacio also got two complaints.
It is time to make dumplings! Ignacio wants to help Ms. Ley with the dough. He plans to roll it out and cut it into small squares.
Ignacio was asked to peel all the heads of garlic in the crate.
Associative Property of Multiplication
Commutative Property of Multiplication
Associative Property of Multiplication
am⋅an=am+n
Multiply
We are told that 1 gigabyte is equivalent to 2^(30) bytes. Multiply this number by 2 to find how many bytes are in 2 gigabytes. 2^(30) * 2 All of the answer options are powers of 2, so let's write our answer the same way. We can use the Product of Powers Property to do this. First, recall that if no power is written on a number, the power is assumed to be 1. Let's can rewrite 2 as 2^1. 2^(30) * 2 = 2^(30) * 2^1 We can now apply the Product of Powers Property. 2^(30) * 2^1 &= 2^(30+ 1) &= 2^(31) We found that there are 2^(31) bytes in 2 gigabytes.
We can find how many bytes there are in 8 gigabytes in a similar fashion. We first need to write 8 as a power of 2.
8= 2^3
There are 2^(30) bytes in 1 gigabyte. Multiply this number by 8=2^3 to find how many bytes there are in 8 gigabytes.
2^(30) * 8 = 2^(30) * 2^3
We can now use the Product of Powers Property.
2^(30) * 2^3 &= 2^(30+ 3)
&= 2^(33)
This means that there are 2^(33) bytes in 8 gigabytes.
We are asked to find how many floppy disks would be required to store one gigabyte of information. We know told that 1 gigabyte is equivalent to 2^(30) bytes. 1gigabyte =2^(30)bytes One floppy disk can hold up to 2^(21) bytes of information. We can divide 1 gigabyte by this amount to calculate how many increments of 2^(21) bytes are in 2^(30) bytes. 1gigabyte/2^(21)bytes = 2^(30)bytes/2^(21)bytes Next, let's use the Quotient of Powers Property to write the expression as a power of 2. 2^(30)/2^(21) &= 2^(30- 21) &= 2^9 We would need 2^9 floppy disks with 2^(21) bytes of memory to store just 1 gigabyte of information! In other words, we would need 512 floppy disks! We have much better storage devices these days that can hold many gigabytes worth of information, so floppy disks have now fallen out of use.
Divide the monomials.
Let's take a look at the given expression. x^6y^3/x^2y We can begin by writing the expression as a product of two divisions. One factor will have the variable x and the other factor the variable y. x^6y^3/x^2y = x^6/x^2 * y^3/y Remember that if no power is written on a variable, the power is assumed to be 1. x^6y^3/x^2y = x^6/x^2 * y^3/y^1 We can now use the Quotient of Powers Property to divide the resulting monomials. Do this by subtracting the power of the denominator from the power of the numerator.
Finally, finish by multiplying the monomials. x^6y^3/x^2y = x^4y^2
Let's take a look at the next expression.
p^7q^5/p^2q^4
Let's follow the same steps we did in Part A. We will first write this expression as a product of two divisions. One factor will have the variable p and the other factor the variable q.
p^7q^5/p^2q^4 = p^7/p^2 * q^5/q^4
We can now use the Quotient of Powers Property to divide each of the resulting monomials.
Recall that any number raised to the power of 1 is equal to itself. p^7q^5/p^2q^4 &= p^5 * q^1 &= p^5 * q Finally, multiply the monomials. p^7q^5/p^2q^4 = p^5q