How do multiplication properties and division rules help you multiply and divide?
2 answers:
The properties of multiplication and division are an easier way when making calculations.
Let's see an example for each one.
Mutiplication:
Suppose we are going to multiply:
It is easier to make the product using the distributive property:
Division:
Suppose we have the following division:
By properties of the division, we know that this division is not defined.
Therefore, when we see a division between 0, we know that the answer is that it is not defined.
Answer:
The rules of multiplication and division help to divide and multiply in a faster way.
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Answer:
77
Step-by-step explanation:
Let's let the unknown number be x.
Supplementary angles add up to a total of 180 degrees. Since we already know one angle is 103, this means that:
Subtract 103 from both sides:
So, the measure of its supplementary angle is 77 degrees.
And we are done!
Rectangles are similar figures, thus if scaled copies of each other then the ratios of corresponding sides must be equal
compare ratios of lengths and widths
rectangles A and B
k = =
← ratio of lengths
k = =
← ratio of widths
scale factors are equivalent, hence rectangle A is a scaled copy of B
rectangles C and B
k = =
← ratio of lengths
k = =
← ratio of width
scale factors (k ) are not equal, hence C is not a scaled copy of B
rectangles A and C
k = =
← ratio of lengths
k = ← ratio of widths
the scale factors are not equal hence A is not a scaled copy of C
Notation
I imagine that the expression you are asked to work with is:
When you use a keyboard it is customary to use "^" to denote an exponent is coming so you could have written: 3x^3y+15xy-9x^2y-45y just to be clear.
PART A
To factor out the GCF we are looking for the greatest factor among the terms. Looking at the coefficients (the numbers) the largest number they can all be divided by is 3 so we will pull out a 3. Notice also that each term has a y in it so we can pull out that.
This gives us:
To factor is to write as a product (something times something else). It undoes multiplication so in this case if you take what we got and multiplied it back you should get the expression we started with.
PART B
Start with the answer in part A. Namely,
. For now let's focus only on what is in the parenthesis. We have four terms so let's take them two at a time. I am separating the expression in two using square brackets. ![[( x^{3}+5x)]-[3 x^{2} -15]](https://tex.z-dn.net/?f=%5B%28%20x%5E%7B3%7D%2B5x%29%5D-%5B3%20x%5E%7B2%7D%20-15%5D)
Let's next factor what is in each bracket:
![[( x^{3}+5x)]-[3 x^{2} -15] = [x( x^{2} +5)]-[3( x^{2} +5)]](https://tex.z-dn.net/?f=%5B%28%20x%5E%7B3%7D%2B5x%29%5D-%5B3%20x%5E%7B2%7D%20-15%5D%20%3D%20%5Bx%28%20x%5E%7B2%7D%20%2B5%29%5D-%5B3%28%20x%5E%7B2%7D%20%2B5%29%5D)
Notice that both brackets have the same expression in them so now we factor that out:![[x( x^{2} +5)]-[3( x^{2} +5)] = (x-3)( x^{2} +5)](https://tex.z-dn.net/?f=%20%5Bx%28%20x%5E%7B2%7D%20%2B5%29%5D-%5B3%28%20x%5E%7B2%7D%20%2B5%29%5D%20%3D%20%28x-3%29%28%20x%5E%7B2%7D%20%2B5%29)
Our original expression (the one we started the problem with) had a 3y we already pulled out. We need to include that in the completely factored expression. Doing so we get:
I imagine that the expression you are asked to work with is:
When you use a keyboard it is customary to use "^" to denote an exponent is coming so you could have written: 3x^3y+15xy-9x^2y-45y just to be clear.
PART A
To factor out the GCF we are looking for the greatest factor among the terms. Looking at the coefficients (the numbers) the largest number they can all be divided by is 3 so we will pull out a 3. Notice also that each term has a y in it so we can pull out that.
This gives us:
To factor is to write as a product (something times something else). It undoes multiplication so in this case if you take what we got and multiplied it back you should get the expression we started with.
PART B
Start with the answer in part A. Namely,
Let's next factor what is in each bracket:
Notice that both brackets have the same expression in them so now we factor that out:
Our original expression (the one we started the problem with) had a 3y we already pulled out. We need to include that in the completely factored expression. Doing so we get: