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OleMash [197]
3 years ago
10

Noah has a total of 47 video games. He only buys action and sports games. He has 21 more action games than sport games. How many

action games and how many sports games does he have?
I need 2 equations for this but what are they and how do you solve them?
Mathematics
1 answer:
Zanzabum3 years ago
8 0
A-action games
s-sports games

a+s=47
a=s+21

I replace the action games in the first equation

s+21+s=47
2s+21=47
2s=47 - 21
2s=26
s=26:2
s=13 number of sport games
a=13+21
a=34 number of action games
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Vsevolod [243]

\huge \boxed{\mathfrak{Answer} \downarrow}

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\large \bf \: Re(\frac{\left(5-3i\right)\left(-2+9i\right)}{\left(-2-9i\right)\left(-2+9i\right)})  \\

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\large \bf \: Re(\frac{\left(5-3i\right)\left(-2+9i\right)}{\left(-2\right)^{2}-9^{2}i^{2}})  \\

By definition, i² is -1. Calculate the denominator.

\large \bf \: Re(\frac{\left(5-3i\right)\left(-2+9i\right)}{85})  \\

Multiply complex numbers 5-3i and -2+9i in the same way as you multiply binomials.

\large \bf \: Re(\frac{5\left(-2\right)+5\times \left(9i\right)-3i\left(-2\right)-3\times 9i^{2}}{85})  \\

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Combine the real and imaginary parts in -10+45i+6i+27.

\large \bf \: Re(\frac{-10+27+\left(45+6\right)i}{85})  \\

Do the additions in \sf-10+27+\left(45+6\right)i.

\large \bf Re(\frac{17+51i}{85})  \\

Divide 17+51i by 85 to get \sf\frac{1}{5}+\frac{3}{5}i \\.

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3 0
2 years ago
Match the features of the graph of the rational function.
Sunny_sXe [5.5K]

After applying <em>algebraic</em> analysis we find the <em>right</em> choices for each case, all of which cannot be presented herein due to <em>length</em> restrictions. Please read explanation below.

<h3>How to analyze rational functions</h3>

In this problem we have a rational function, whose features can be inferred by algebraic handling:

Holes - x-values that do not belong to the domain of the <em>rational</em> function:

x³ + 8 · x² - 9 · x = 0

x · (x² + 8 · x - 9) = 0

x · (x + 9) · (x - 1) = 0

x = 0 ∨ x = - 9 ∨ x = 1

But one root is an evitable discontinuity as:

y = (9 · x² + 81 · x)/(x³ + 8 · x² - 9 · x)

y = (9 · x + 81)/(x² + 8 · x - 9)

Thus, there are only two holes. (x = - 9 ∨ x = 1) Besides, there is no hole where the y-intercept should be.

Vertical asymptotes - There is a <em>vertical</em> asymptote where a hole exists. Hence, the function has two vertical asymptotes.

Horizontal asymptotes - <em>Horizontal</em> asymptote exists and represents the <em>end</em> behavior of the function if and only if the grade of the numerator is not greater than the grade of the denominator. If possible, this assymptote is found by this limit:

y = \lim_{x \to \pm \infty} \frac {9\cdot x + 81}{x^{2}+8\cdot x - 9}

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The function has a horizontal asymptote.

x-Intercept - There is an x-intercept for all x-value such that numerator is equal to zero:

9 · x + 81 = 0

x = - 9

There is a x-intercept.

Lastly, we have the following conclusions:

  1. How many holes? 2
  2. One <em>horizontal</em> asymptote along the line where y always equals what number: 0
  3. This function has x-intercepts? True
  4. One <em>vertical</em> asymptote along the line where x always equals what number: 1
  5. There is a hole where the y-intercept should be? False

To learn more on rational functions: brainly.com/question/27914791

#SPJ1

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sp2606 [1]

Angle BCA

Step-by-step explanation:

You can see this due to the angle having the name amount of congruent angle marks.

5 0
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