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3 years ago

Break apart 372*7 multiplication

2 answers:

7 0

372 simplified is: 2*2*3*31. Thus, 372*7= 2*2*3*31*7. I believe that's what the question is asking for, Hope that helps. :)

tigry1 [53]3 years ago

3 0

372 is simplified 2*2*3*31*

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John has 9 different flags that he wants to hang on the wall. How many different ways can the flags be arranged in a row?

Masteriza [31]

Answer:

answer is 362,880

Step-by-step explanation:

9!= 9×8×7×6×5×4×3×2×1=362,880

4 0

4 years ago

Read 2 more answers

CAN SOMEBODY HELP ME PLEASE

Mashutka [201]

Answer:

The answer to your questions are:

15.- 45 mg

16.- 25 kg

17.- 185 pounds

18.- 6 mg

Step-by-step explanation:

D = (a M) / (a + 12)

15.-

a = 4 years

M = 180

D = (4 x 180) / (4 + 12) = 720 / 16

D = 45 mg

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55 pounds ------------ x

x = 55 x 0.4535 / 1

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M = 18/3 = 6 mg

5 0

3 years ago

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The graph of f(x) = |x| is transformed to g(x) = |x + 1| – 7. On which interval is the function decreasing?

timofeeve [1]

Answer:

g is decreasing for all x less than or equal to -1

Step-by-step explanation:

The graph of the function y = |x| looks like a "vee."

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Function g(x) is decreasing for all $x \le -1$ .

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3 years ago

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olya-2409 [2.1K]

Answer:

The option which you chose is absolutely right.

Step-by-step explanation:

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3 years ago

Select either relation (if the set is a relation but not a function), function (if the set is both a relation and a function), o

galina1969 [7]

Answer: The correct options are

(A) relation

(B) function.

Step-by-step explanation: We are given to check whether the following set is a relation or a relation and a function or neither :

A = {(1, 2) (2, 2) (3, 2) (4, 2)} .

Let us consider the following two sets :

U = {1, 2, 3, 4} and V = {2}.

Then, we see that the Cartesian Product of U and V is given by

U × V = {(1, 2) (2, 2) (3, 2) (4, 2)}.

Since a relation is a subset of Cartesian product, so the given set A is a relation.

Also, we know that a relation is a function if every first element is associated with one second element.

Since the relation A satisfies this condition, so the relation A is a function.

Thus, the given set A is both a relation and a function.

Option (A) and (B) are correct.

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