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tatuchka [14]
3 years ago
8

Alisa says it is easier to compare the numbers in set a than in set b.

Mathematics
2 answers:
Sidana [21]3 years ago
8 0
I don't know vrghhhhhhhhhhhh
Oksana_A [137]3 years ago
7 0

Answer:

She can use the knowledge of place value to compare the numbers in set a than in set b.

Step-by-step explanation:

Consider the provided sets.

According to place value chart:

  Millions    Hundred Th.  Ten Th.    Thousands  hundreds Tens Ones

1,000,000     100,000      10,000       1000             100          10       1

Consider the set A) 45,760 & 1,025,680

In number 45,760 the first digit from left is 4 which is at ten thousands place.

In number 1,025,680 the first digit from left is 1 which is at million place.

So, it is easy to identify which number is greater by just seeing the greatest place value of the number.

Consider the set B. 492,111 & 409,867

Now consider the number 492,111 and 409,867

In number 492,111 the first digit from left is 4 which is at hundred thousands place.

In number 409,867 the first digit from left is 4 which also at hundred thousands place.

The greatest place value of both the number is same, so by seeing greatest place value number we can not identify which number is greater, to identify which number is greater we further need to compare the digit at ten thousands place.

Hence, by using the knowledge of place value she can compare the numbers in set a than in set b.

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\text{Hi there! :D}

\boxed{\text{There is a horizontal shift of 1 unit left and 5 units down.}}

\text{Recall that the transformations form of a quadratic function is:}\\\\y = \pm a(b(x-h))+k \text{ where:}\\\\a = \text{ vertical stretch/compression}\\\\b = \text{ horizontal stretch/compression}\\\\h = \text{ shift left/right}\\\\k = \text{ shift up/down}\\\\\text{Therefore, in this instance:}\\\\h = -1\\\\k = -5, \text{ so:}\\\\\text{There is a horizontal shift of 1 unit left and 5 units down.}

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