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hammer [34]
3 years ago
15

Suppose 60 cars start at a car race. in how many ways can the top 3 cars finish the​ race?

Mathematics
1 answer:
lapo4ka [179]3 years ago
5 0
60 different cars could finish first place
of the 59 that didn't win, any of the 59 could.
finish second
of the remaining 58, any could finish 3rd, or 58
possibilities.

This our answer is 60*59*58=205,320
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Tomtit [17]

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8 0
3 years ago
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Vikentia [17]

Answer:

3x^2 + 3  -->  6, 15, 30, 51

2x^2 - 1    -->  1, 7, 17, 31

x^2 + 2     -->  3, 6, 11, 18

Step-by-step explanation:

Let's start with the first equation; 3x^2 + 3

Substitute 1 (the first digit in a sequence) for x.

3(1^2) + 3

3(1) + 3

3 + 3 = <u>6</u>

3(2^2) + 3 Then the second digit.

3(4) + 3

12 + 3 = <u>15</u>

Since the two numbers we have so far are 6 and 15, there is only one sequence this could match. 6, 15, 30, 51.

2(1^2) - 1

2(1) - 1

2 - 1 = <u>1</u>

This equation represents 1, 7, 17, 31.

These same steps apply to the other equation as well.

1^2 + 2, then 2^2 + 2, then 2^2 + 2, and so on. (But we don't need to do extra work to figure that out.)

7 0
2 years ago
11xto the 7th power ×6x to the 5th power
solong [7]

The first one is 19,487,171 and the second is 7,776

6 0
3 years ago
Tell the error and leave if you disagree or agree.​
Musya8 [376]

Answer:

Step-by-step explanation:

9. disagree;

error: he should have divided 60 by 4, not subtracted. So the correct answer is x = 15.

10. disagree;

error: it should be 3x + 2 = 127 (opposite angles)

so x = 125/3

8 0
3 years ago
Evaluate the expression. (6-2)[(2+3)+4<br> A) 9 <br> B) 13 <br> C) 24<br> D)36
Kipish [7]
(6-2)(2+3)+4

(4)(5)+4

20+4

=24
7 0
3 years ago
Read 2 more answers
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