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Taya2010 [7]
3 years ago
8

You have 4 reindeer, Bloopin, Balthazar, Gloopin, and Prancer, and you want to have 3 fly your sleigh. You always have your rein

deer fly in a single-file line
Mathematics
1 answer:
Setler79 [48]3 years ago
8 0

Answer:

P43=4!(4–3)!=241=24 24 possible choices

Step-by-step explanation:

There are four choices you can make for the lead reindeer. For each possible choice, there are then three remaining you can choose to fly second, making 4×3=12 choices for the lead pair. For each possible choice there are two remaining reindeer to take up the back position, making 12×2=24 choices for the team of three.

This type of problem is called a permutation problem, and we write Pnr for the number of ways of choosing r items from n possibilities when the order of the items matters. In this case we are choosing 3 reindeer from 4 possibilities, and the order they appear in the flying line does matter, so the answer we want is P43. The general formula is Pnr=n!(n−r)!. For the answer we are looking for we therefore have:

P43=4!(4–3)!=241=24

Brainliest?

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Nadusha1986 [10]
\rm \int \dfrac{x^2+7}{x^2+2x+5}~dx

Derivative of the denominator:
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Hmm our numerator is 2x+7. Ok this let's us know that a simple u-substitution is NOT going to work. But let's apply some clever Algebra to the numerator splitting it up into two separate fractions. Split the +7 into +2 and +5.

\rm \int \dfrac{x^2+2+5}{x^2+2x+5}~dx

and then split the fraction,

\rm \int \dfrac{x^2+2}{x^2+2x+5}~dx+\int\dfrac{5}{x^2+2x+5}~dx

Based on our previous test, we know that a simple substitution will work for the first integral: \rm \quad u=x^2+2x+5\qquad\to\qquad du=2x+2~dx

So the first integral changes,

\rm \int \dfrac{1}{u}~du+\int\dfrac{5}{x^2+2x+5}~dx

integrating to a log,

\rm ln|x^2+2x+5|+\int\dfrac{5}{x^2+2x+5}~dx

Other one is a little tricky. We'll need to complete the square on the denominator. After that it will look very similar to our arctangent integral so perhaps we can just match it up to the identity.

\rm x^2+2x+5=(x^2+2x+1)+4=(x+1)^2+2^2

So we have this going on,

\rm ln|x^2+2x+5|+\int\dfrac{5}{(x+1)^2+2^2}~dx

Let's factor the 5 out of the intergral,
and the 4 from the denominator,

\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\frac{(x+1)^2}{2^2}+1}~dx

Bringing all that stuff together as a single square,

\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(\dfrac{x+1}{2}\right)^2+1}~dx

Making the substitution: \rm \quad u=\dfrac{x+1}{2}\qquad\to\qquad 2du=dx

giving us,

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simplying a lil bit,

\rm ln|x^2+2x+5|+\frac52\int\dfrac{1}{u^2+1}~du

and hopefully from this point you recognize your arctangent integral,

\rm ln|x^2+2x+5|+\frac52arctan(u)

undo your substitution as a final step,
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Hope that helps!
Lemme know if any steps were too confusing.

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