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miskamm [114]
3 years ago
10

Show how to solve the problem 378x6 using place value with regrouping

Mathematics
1 answer:
Ymorist [56]3 years ago
7 0
First you arrange it vertically then multiply the answer is 2,268
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Anna is looking for a new part-time job. They respond to a classified ad for
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Answer: $2,393.42

Step-by-step explanation:

I suppose that the salary is yearly.

So in one year, 365 days, the net pay would be $62,400.

Now, "by-weekly" refers to two weeks or 14 days.

If she gets paid $62,400 in 365 days.

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N = (14/365)*$62,400 = $2,393.42

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3 years ago
Using the drawing, what is the vertex of angle 4?
FinnZ [79.3K]
The vertex is where two lines make a point and the answer would be letter C. A

4 0
3 years ago
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Please help i dont understand this
zavuch27 [327]

Answer:

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Step-by-step explanation:

4 0
3 years ago
Three stamps can be attached to each other in various ways.how many ways might three stamps be attached?
ivanzaharov [21]
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.


8 0
3 years ago
multiply the polynomials. What is the coefficient of the last term? −4m(2m3 −6m2 + m − 5) = −8m4 +24m3 −4m2 ___m
lesya692 [45]
+20m . Hope i helped
6 0
2 years ago
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