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Paladinen [302]
3 years ago
7

Add: (4a.”+ 6a2- 10a - 4) +(6a + a²+ a + 7)

Mathematics
1 answer:
sineoko [7]3 years ago
3 0
(2a-1 )exponent two•(3a + 2) exponent 2• (4a-3) exponent 2
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A bacteria culture begins with 4 bacteria which double in size every hour. How many bacteria exist in the culture after 8 hours?
Shalnov [3]

Answer:

Option C. 1,024 bacteria

Step-by-step explanation:

We are given that a bacteria culture begins with 4 bacteria which double in size every hour. And we have to find that how many bacteria exist in the culture after 8 hours;

So, Number of bacteria after one hour = 4 * 2 = 8

     Number of bacteria after two hours = 8 * 2 = 16

     Number of bacteria after three hours = 16 * 2 = 32

     Number of bacteria after four hours = 32 * 2 = 64

     Number of bacteria after five hours = 64 * 2 = 128

     Number of bacteria after six hours = 128 * 2 = 256

     Number of bacteria after seven hours = 256 * 2 = 512

     Number of bacteria after eight hours = 512 * 2 = 1,024

Therefore, 1,024 bacteria exist in the culture after 8 hours .    

7 0
3 years ago
Yo I’m I’m stuck on this. <br> can I get some help pls
ikadub [295]

Answer:

Which problem?

Step-by-step explanation:

7 0
3 years ago
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What does y= -3x+3 equal
Andre45 [30]

Answer:

Solving for x, the answer is x = 1 - y/3

Hope this helps :D

7 0
3 years ago
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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
Find the area (pls help)
icang [17]

Answer:

15

Step-by-step explanation:

7 0
3 years ago
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