Answer:
B. 21 and 19/25
Step-by-step explanation:
Rewrite the decimal number as a fraction with 1 in the denominator
21.76=21.761
Multiply to remove 2 decimal places. Here, you multiply top and bottom by 102 = 100
21.761×100100=2176100
Find the Greatest Common Factor (GCF) of 2176 and 100, if it exists, and reduce the fraction by dividing both numerator and denominator by GCF = 4,
2176÷4100÷4=54425
Simplify the improper fraction,
=211925
In conclusion,
21.76=211925
Answer:
The relationship between the angles is alternate exterior angles.
Step-by-step explanation:
Exterior means outside.
Alternate means every other or in this case across form each other.
These angles represent alternate exterior angles because they are on the outside of the transversal and they are alternating each other.
10m + 80 = P
where every month he gets 10, and for 6 months, he now has 60
plus the 80 he already has
so (10 * 6 ) + 80 = 140
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.
Answer:
8
Step-by-step explanation:
