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Igoryamba
3 years ago
12

How to write 100, 203 in two other forms

Mathematics
1 answer:
RUDIKE [14]3 years ago
7 0
<span>So we are wondering how can we write the number 100203 in two different forms. First form can be word form: one hundred thousand two hundred and three. Second form can be a fraction: 100203/1 or 1002030/10 or 10020300/100 and so on. Third form can be adition expression: 100000 + 200 + 3. </span>
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PLEASE ANSWER AND EXPLAIN 99 POINTS [URGENT]
givi [52]

Answer:

A

Step-by-step explanation:

The farther you move through x, the more y decreases

4 0
3 years ago
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Where is the answer to the expression 3 - 6 located on a horizontal number line? O 3 units to the left of 3 O 3 units to the rig
trapecia [35]

Answer:

3

Step-by-step explanation:

because if you are doing negatives it is kinda like subtracting

i \: hope \: this \: helps \\

6 0
3 years ago
If (-9, 2) is on the graph of Flx), which point must be on the graph of the
dmitriy555 [2]

Answer:

(2, - 9 )

Step-by-step explanation:

Since 2 is the output when - 9 is input to f(x)

Then reversing the procedure, that is the inverse gives an output of - 9 for an input of 2.

(- 9, 2) is a point on the graph of f(x), then

(2, - 9) is a point on the graph o the inverse function f^{-1}(x)

7 0
3 years ago
When figuring out the amount of mulch would be needed for Alex's back yard, he created an equation that
soldier1979 [14.2K]

Answer:

Alex will need 14 bags

Step-by-step explanation:

B= 4(2x+7)+3(2x+7)

and

2x+7=2

First, we should solve the second equation to obtain x:

2x+7=2

2x=2-7

x= -5/2

Then, we use this answer to calculate B:

B=4(2(-5/2)+7)+3(2)

B=4(-5+7)+6

B=4(2)+6

B=8+6

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Alex will need 14 bags

7 0
3 years ago
Eights rooks are placed randomly on a chess board. What is the probability that none of the rooks can capture any of the other r
erastova [34]

Answer:

The probability is \frac{56!}{64!}

Step-by-step explanation:

We can divide the amount of favourable cases by the total amount of cases.

The total amount of cases is the total amount of ways to put 8 rooks on a chessboard. Since a chessboard has 64 squares, this number is the combinatorial number of 64 with 8, 64 \choose 8 .

For a favourable case, you need one rook on each column, and for each column the correspondent rook should be in a diferent row than the rest of the rooks. A favourable case can be represented by a bijective function  f : A \rightarrow A , with A = {1,2,3,4,5,6,7,8}. f(i) = j represents that the rook located in the column i is located in the row j.

Thus, the total of favourable cases is equal to the total amount of bijective functions between a set of 8 elements. This amount is 8!, because we have 8 possibilities for the first column, 7 for the second one, 6 on the third one, and so on.

We can conclude that the probability for 8 rooks not being able to capture themselves is

\frac{8!}{64 \choose 8} = \frac{8!}{\frac{64!}{8!56!}} = \frac{56!}{64!}

7 0
3 years ago
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