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olga nikolaevna [1]
3 years ago
10

Jason went to a carnival where he could goon all rides for a flat fee of $30 but he had to pay $2 for each arcade game he played

. Jason spent $44. How many arcade games did he play?
Mathematics
1 answer:
enot [183]3 years ago
4 0

Answer:

He played 7 arcade games.

Step-by-step explanation:

The amount paid in relation to the number of games played can be modeled by a linear function in the following format:

A(n) = F + gn

In which F is the flat rate and g is the price of each game.

Flat fee of $30 but he had to pay $2 for each arcade game he played.

This means that F = 30, g = 2

So

A(n) = 30 + 2n

Jason spent $44. How many arcade games did he play?

This is n for which A(n) = 44. So

A(n) = 30 + 2n

44 = 30 + 2n

2n = 14

n = \frac{14}{2}

n = 7

He played 7 arcade games.

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What is the area of the figure down below
kogti [31]

68


Explanation:

So to get this answer you need to split it into smaller shapes. We already have the squares so we know that 3*3 is 9 and then we double that because there are two of them.

So, so far we have 18+x

We need to take the middle shape and split it into two trapezoids.

The area for trapezoids is T=(a+b/2)h

So T= (3+7/2)5
T= 5*5=25
And then we would double that because there are two of them so 50 and then fill it in for x
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The radius of circle C is 6 units and the measure of central angle ACB is I radians. What is the approximate area of the entire
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Step-by-step explanation:

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X squared+ y squared = 2 y = 2x squared – 3 Which of the following describes the system?
cluponka [151]

Answer:

x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}} and y=-1,\frac{1}{2}

The ordered pair solutions are (-\sqrt{\frac{7}{4}},0.5), (\sqrt{\frac{7}{4}},0.5), (-1,-1), and (1,-1).

Step-by-step explanation:

I'm assuming the system is \left \{ {x^2+y^2=2} \atop {y=2x^2-3}} \right.:

x^2+y^2=2

x^2+(2x^2-3)^2=2

x^2+(4x^4-12x^2+9)=2

x^2+4x^4-12x^2+9=2

4x^4-11x^2+9=2

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x^4-11x^2+28=0

(x^2-7)(x^2-4)=0

(4x^2-7)(x^2-1)=0

4x^2-7=0

4x^2=7

x^2=\frac{7}{4}

x=\pm\sqrt{\frac{7}{4}}

x^2-1=0

x^2=1

x=\pm1

y=2x^2-3

y=2(\pm\sqrt{\frac{7}{4}})^2-3

y=2({\frac{7}{4}})-3

y=\frac{7}{2}-3

y=\frac{1}{2}

y=2x^2-3

y=2(\pm1)^2-3

y=2(1)-3

y=2-3

y=-1

Therefore, x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}} and y=-1,\frac{1}{2}

The ordered pair solutions are (-\sqrt{\frac{7}{4}},0.5), (\sqrt{\frac{7}{4}},0.5), (-1,-1), and (1,-1).

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