Answer: 123
Step-by-step explanation:
The ratio of of number of homework papers to number of exit tickets of Mr Rowley and Ms. Alvera are not equivalent.
<h3>Ratio</h3>
A ratio is a number representing a comparison between two named things. It is also the relative magnitudes of two quantities usually expressed as a quotient.
Mr Rowley:
- Homework papers = 16
- Tickets to return = 2
Ratio of number of homework papers to number of exit tickets = 16 : 2
= 16 / 2
= 8 / 1
= 8 : 1
Ms Alvera:
- Homework papers = 64
- Tickets to return = 60
Ratio of number of homework papers to number of exit tickets = 64 : 60
= 64/60
= 16 / 15
= 16 : 15
Therefore, the ratio of of number of homework papers to number of exit tickets of Mr Rowley and Ms. Alvera are not equivalent.
Learn more about ratio:
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The correct answer is C) (5m^50 - 11n^8) (5m^50 + 11n^8)
We can tell this because of the rule regarding factoring the difference of two perfect squares. When we have two squares being multiplied, we can use the following rule.
a^2 - b^2 = (a - b)(a + b)
In this case, or first term is 25m^100. So we can solve that by setting it equal to a^2.
a^2 = 25m^100 -----> take the square root of both sides
a = 5m^50
Then we can do the same for the b term.
b^2 = 121n^16 ----->take the square root of both sides
b = 11n^8
Now we can use both in the equation already given
(a - b)(a + b)
(5m^50 - 11n^16)(5m^50 + 11n^16)
Answer:
(-9,-1)
Step-by-step explanation:
You plug in the x and y values and solve the algebraic expression.
Let's call the stamps A, B, and C. They can each be used only once. I assume all 3 must be used in each possible arrangement.
There are two ways to solve this. We can list each possible arrangement of stamps, or we can plug in the numbers to a formula.
Let's find all possible arrangements first. We can easily start spouting out possible arrangements of the 3 stamps, but to make sure we find them all, let's go in alphabetical order. First, let's look at the arrangements that start with A:
ABC
ACB
There are no other ways to arrange 3 stamps with the first stamp being A. Let's look at the ways to arrange them starting with B:
BAC
BCA
Try finding the arrangements that start with C:
C_ _
C_ _
Or we can try a little formula; y×(y-1)×(y-2)×(y-3)...until the (y-x) = 1 where y=the number of items.
In this case there are 3 stamps, so y=3, and the formula looks like this: 3×(3-1)×(3-2).
Confused? Let me explain why it works.
There are 3 possibilities for the first stamp: A, B, or C.
There are 2 possibilities for the second space: The two stamps that are not in the first space.
There is 1 possibility for the third space: the stamp not used in the first or second space.
So the number of possibilities, in this case, is 3×2×1.
We can see that the number of ways that 3 stamps can be attached is the same regardless of method used.
