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Strike441 [17]
3 years ago
10

Any one know the answer

Mathematics
2 answers:
Komok [63]3 years ago
7 0

answer: b

work:

let's solve by looking at the options.

<em>option one: </em>we can see right away that this is incorrect because it involves multiplication, and the first operation is multiplying 30 x 10. since the number is 30.425, only slightly more than 30, we know this isn't an option since 30 x 10 is 300, and the number only gets bigger from there.

<em>option two: </em>this option is correct. in standard form, thirty is 30, and four hundred twenty five thousandths is ".425", so putting those together you get 30.425

<em>options three and four:  </em>we already know these are incorrect because b is the answer, but option three has adding 4 to 3 x 10, which results in 34, which is already larger than 30.425. lastly, the last one is also incorrect because it doesn't have thousandths at the end, which it needs, and doesn't make sense otherwise.

Mice21 [21]3 years ago
4 0

Im pretty sure it is B

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Hi again here's the picture
pshichka [43]

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<h2>351.88 m</h2>

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Alice and Bob play a game involving a circle whose circumference is divided by 12 equally-spaced points. The points are numbered
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Answer:

Step-by-step explanation:

2005 AMC 8 Problems/Problem 20

Problem

Alice and Bob play a game involving a circle whose circumference is divided by 12 equally-spaced points. The points are numbered clockwise, from 1 to 12. Both start on point 12. Alice moves clockwise and Bob, counterclockwise. In a turn of the game, Alice moves 5 points clockwise and Bob moves 9 points counterclockwise. The game ends when they stop on the same point. How many turns will this take?

$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 24$

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See Also

2005 AMC 8 (Problems • Answer Key • Resources)

Preceded by

Problem 19 Followed by

Problem 21

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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