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crimeas [40]
3 years ago
8

A video game has the following formula: For every 1,333 in attack power, you gain 3 levels. You get one bonus star per every 38

levels completed. If a player has an attack power of 0.21 million, how many bonus stars could they get? Round to a whole number.
Mathematics
1 answer:
aleksley [76]3 years ago
7 0

Answer:

Rounded to a whole number, the player has 12 bonus stars.

Step-by-step explanation:

Since the video game establishes that for every 1,333 points of attack power 3 levels are gained, and that the player has a total of 0.21 million attack points, to determine the number of levels that player has it is necessary to perform the following calculation :

0.21 million = 210,000

210,000 / 1,333 x 3 = Levels

472.6 = Levels

Now, every 38 levels, the game awards a bonus star. To determine the amount of bonus stars that the player has, the following division must be made:

472.6 / 38 = X

12.43 = X

Thus, rounded to a whole number, the player has 12 bonus stars.

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For part (a), you have

\dfrac x{x^2+x-6}=\dfrac x{(x+3)(x-2)}=\dfrac a{x+3}+\dfrac b{x-2}
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If x=2, then 2=b(2-3)\implies b=-2.

If x=-3, then -3=a(-3-2)\implies a=\dfrac35.

So,

\dfrac x{x^2+x-6}=\dfrac 3{5(x+3)}-\dfrac 2{x-2}

For part (b), since the degrees of the numerator and denominator are the same, you first need to find the quotient and remainder upon division.

\dfrac{x^2}{x^2+x+2}=\dfrac{x^2+x+2-x-2}{x^2+x+2}=1-\dfrac{x+2}{x^2+x+2}

In the remainder term, the denominator x^2+x+2 can't be factorized into linear components with real coefficients, since the discriminant is negative (1-4\times1\times2=-7). However, you can still factorized over the complex numbers, so a partial fraction decomposition in terms of complexes does exist.

x^2+x+2=0\implies x=-\dfrac12\pm\dfrac{\sqrt7}2i
\implies x^2+x+2=\left(x-\left(-\dfrac12+\dfrac{\sqrt7}2i\right)\right)\left(x-\left(-\dfrac12-\dfrac{\sqrt7}2i\right)\right)
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Then you have

\dfrac{x+2}{x^2+x+2}=\dfrac a{x+\dfrac12-\dfrac{\sqrt7}2i}+\dfrac b{x+\dfrac12+\dfrac{\sqrt7}2i}
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When x=-\dfrac12-\dfrac{\sqrt7}2i, you have

-\dfrac12-\dfrac{\sqrt7}2i+2=b\left(-\dfrac12-\dfrac{\sqrt7}2i+\dfrac12-\dfrac{\sqrt7}2i\right)
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When x=-\dfrac12+\dfrac{\sqrt7}2i, you have

-\dfrac12+\dfrac{\sqrt7}2i+2=a\left(-\dfrac12+\dfrac{\sqrt7}2i+\dfrac12+\dfrac{\sqrt7}2i\right)
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So, you could write

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but that may or may not be considered acceptable by that webpage.
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