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Schach [20]
3 years ago
14

Ten runners join a race. In how many possible ways can they be arranged as first, second, and third placer? Pls answer

Mathematics
2 answers:
dybincka [34]3 years ago
7 0
Ans: 720

<span>This is equal to the number of different arrangements of three persons from a group of 10 persons </span>

Required number of ways = 10P3 = 720

OR can be solved in this way
select 3 runners (10C3 ways)
These 3 runners can be arranged in 3! ways
<span>Total = 10C3×3! = 720 .hope i was able to help :)</span>
Daniel [21]3 years ago
5 0
Ans: 720

<span>This is equal to the number of different arrangements of three persons from a group of 10 persons </span>

Required number of ways = 10P3 = 720

OR can be solved in this way
select 3 runners (10C3 ways)
These 3 runners can be arranged in 3! ways
<span>Total = 10C3×3! = 720

</span>
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Serjik [45]

The answer is C. -4n⁴+3n.


Separate the expression by its like terms.

12n-2n⁴-10n-2n⁴+n

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The simplified expression comes out to -4n⁴+3n.

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Find the width of a rectangle 25 feet and the perimeter is 80​
lukranit [14]

\huge\underline{\red{A}\blue{n}\pink{s}\purple{w}\orange{e}\green{r} -}

  • Given - <u>a </u><u>rectangle </u><u>with </u><u>length</u><u> </u><u>2</u><u>5</u><u> </u><u>feet </u><u>and </u><u>perimeter </u><u>8</u><u>0</u><u> </u><u>feet</u>

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We know that ,

\bold{Perimeter \: of \: rectangle = 2(l + b)} \\

where <u>b </u><u>=</u><u> </u><u>width </u><u>/</u><u> </u><u>breadth</u> of rectangle

‎ ‎ ‎

<u>substituting</u><u> </u><u>the </u><u>values </u><u>in </u><u>the </u><u>formula </u><u>stated </u><u>above </u><u>,</u>

\bold{80  = 2(25 + b)} \\  \\\bold{ \implies \: 25 + b = \cancel \frac{80}{2} } \\  \\ \bold{\implies \: 25 + b = 40 }\\  \\ \bold{\implies \: b = 40 - 25 }\\  \\\bold{ \implies \: b = 15 \: feet}

hope helpful ~

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