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Vadim26 [7]
3 years ago
11

A tennis coach took his team out for lunch and bought 8 hamburgers and 5 fries for $24. The players were still hungry so the coa

ch bought six more hamburgers and two more fries for $16.60. find the cost of each.
Mathematics
1 answer:
g100num [7]3 years ago
5 0

The cost of 1 hamburger is $ 2.5 and cost of 1 fries is $ 0.8

<h3><u>Solution:</u></h3>

Let "f" be the cost of 1 fries

Let "h" be the cost of 1 hamburger

<em><u>Given that, tennis coach took his team out for lunch and bought 8 hamburgers and 5 fries for $24</u></em>

8 x cost of 1 hamburger + 5 x cost of 1 fries = 24

8 \times h + 5 \times f = 24

8h + 5f = 24 -------- eqn 1

<em><u>The players were still hungry so the coach bought six more hamburgers and two more fries for $16.60</u></em>

6 x cost of 1 hamburger + 2 x cost of 1 fries = 16.60

6 \times h + 2 \times f = 16.60

6h + 2f = 16.60 ------ eqn 2

<em><u>Let us solve eqn 1 and eqn 2</u></em>

Multiply eqn 1 by 2

16h + 10f = 48 ------ eqn 3

Multiply eqn 2 by 5

30h + 10f = 83 -------- eqn 4

<em><u>Subtract eqn 3 from eqn 4</u></em>

30h + 10f = 83

16h + 10f = 48

( - ) ----------------------

14h = 35

<h3>h = 2.5</h3>

Substitute h = 2.5 in eqn 1

8(2.5) + 5f = 24

20 + 5f = 24

5f = 4

<h3>f = 0.8</h3>

Thus cost of 1 hamburger is $ 2.5 and cost of 1 fries is $ 0.8

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Step-by-step explanation:

The equation of a circle in the center-radius form is:

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Where (h,k) are the coordinates of the center and r is the radius.

Now, we are given the equation of this circle as follows:

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Rearranging the equation:

x^{2}+6x+y^{2}-8y=75 (4)

(x^{2}+6x)+(y^{2}-8y)=75 (5)

Now we have to complete the square in both parenthesis, in order to have a perfect square trinomial in the form of (a\pm b)^{2}=a^{2}\pm+2ab+b^{2}:

<u>For the first parenthesis:</u>

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We can rewrite this as:

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x^{2}+2(3)x+3^{2}=x^{2}+6x+9=(x+3)^{2}

<u>For the second parenthesis:</u>

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We can rewrite this as:

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Hence in this case b=-3 and b^{2}=9:

y^{2}-2(4)y+4^{2}=y^{2}-8y+16=(y-4)^{2}

Then, equation (5) is rewritten as follows:

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<u>Note we are adding 9 and 16 in both sides of the equation in order to keep the equality.</u>

Rearranging:

(x-3)^{2}+(y-4)^{2}=100 (7)

At this point we have the circle equation in the center radius form (x-h)^{2} +(y-k)^{2}=r^{2}

Hence:

h=-3

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r=\sqrt{100}=10

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