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den301095 [7]
3 years ago
8

*NEED HELP ASAP*

Mathematics
2 answers:
GrogVix [38]3 years ago
7 0
X^2 + y^2 - 2x + 7y + 1 = 0


(x^2 - 2x) + (y^2 + 7y) + 1 = 0


(x^2 - 2x + 1) + (y^2 + 7y) + 1 = 0+1


(x^2 - 2x + 1) + (y^2 + 7y + 49/4) + 1 = 0+1+49/4


(x - 1)^2 + (y + 7/2)^2 + 1 = 0+1+49/4


(x - 1)^2 + (y + 7/2)^2 + 1-1 = 0+1+49/4-1


(x - 1)^2 + (y + 7/2)^2 = 49/4


(x - 1)^2 + (y + 7/2)^2 = (7/2)^2


The final answer is choice B
enot [183]3 years ago
7 0

Answer:

The correct option B (x -1)^{2} + (y +\frac{7}{2})^{2} = (\frac{7}{2})^{2}

Step-by-step explanation:

We need to find out the correct option which is similar to the expression;

x^{2} + y^{2}- 2x + 7y + 1 = 0

combine the similar variable together

(x^{2} - 2x) + (y^{2} + 7y) + 1 = 0

Add 1 both the sides,

(x^{2} - 2x + 1) + (y^{2} + 7y) + 1 = 0+1

Add both the sides by \frac{49}{4}

(x^{2} - 2x + 1) + (y^{2} +\frac{49}{4}+ 7y) + 1 = 0+1 + \frac{49}{4}

(x^{2} - 2x + 1) + (y +\frac{7}{2})^{2} + 1 = 0+1 + \frac{49}{4}

Subtract both the sides by 1,

(x^{2} - 2x + 1) + (y +\frac{7}{2})^{2} + 1-1 = 0+1 + \frac{49}{4}-1

(x^{2} - 2x + 1) + (y +\frac{7}{2})^{2} = \frac{49}{4}

(x -1)^{2} + (y +\frac{7}{2})^{2} = (\frac{7}{2})^{2}

This is equivalent to option B

Therefore the correct option B (x -1)^{2} + (y +\frac{7}{2})^{2} = (\frac{7}{2})^{2}

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Answer:

x = 1178 games

Step-by-step explanation:

Let the number of games = x

Let the total cost = Tc

Let the total revenue = Tr

Given the following data;

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Cost of each game = $1.50

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Total cost, Tc = (Cost of each game * Number of games) + Investment

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1.50x + 10000 = 9.99x

9.99x - 1.50x = 10000

8.49x = 10000

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x = 1177.86 ≈ 1178 games.

<em>Therefore, the number of games that must be sold before the business breaks even is 1178 games. </em>

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