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MissTica
3 years ago
9

Katy's favorite rides at the amusement park are the roller coasters and water slide. The wait time for the roller coaster is 25

minutes and the wait time for the water slide is 10 minutes. If she went on 12 rides total and waited 3 hours in line, how many times did she go on each ride.
Mathematics
1 answer:
ioda3 years ago
7 0

Answer:

She went on the slide 8 times and on the roller coaster 4 times

Step-by-step explanation:

We convert each statements to a mathematical equation.

Firstly, let's represent the number of times she went on the coaster with R and the number of times on the slide with S. We know quite well she went on 12 rides. Hence the summation of both number of times yield 12.

Mathematically, R + S = 12. ........(i)

Now we also know her total wait time was 3hours. Since an hour equals 60 minutes, her total wait time would equal 180 minutes.

To get a mathematical representation for the wait time, we multiply the number of roller coaster rides by 25 and that of the slides by 10.

Mathematically, 25R + 10S = 180 .......(ii)

Here we now have two equations that we can solve simultaneously.

From equation 1 we can say R = 12 - S. We can then substitute this into equation 2 to yield the following:

25(12 - s) + 10s = 180

300 - 25s + 10s = 180

300 - 25s + 10s = 180

300 - 15s = 180

15s = 300 - 180

15s = 120

S = 120/15

S = 8

S = 8 , and R = 12 - S = 12 - 8 = 4

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Evaluate a(b - c ^ 2) * if * a = 2/3, b = 3/4, c = 1/2 A: 1/65 B: 1/3 C: 1/4 D: 2/3
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Answer:

If a+b+c=1,

a

2

+

b

2

+

c

2

=

2

,

a

3

+

b

3

+

c

3

=

3

then find the value of

a

4

+

b

4

+

c

4

=

?

we know

2

(

a

b

+

b

c

+

c

a

)

=

(

a

+

b

+

c

)

2

−

(

a

2

+

b

2

+

c

2

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⇒

2

(

a

b

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b

c

+

c

a

)

=

1

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−

2

=

−

1

⇒

a

b

+

b

c

+

c

a

=

−

1

2

given

a

3

+

b

3

+

c

3

=

3

⇒

a

3

+

b

3

+

c

3

−

3

a

b

c

+

3

a

b

c

=

3

⇒

(

a

+

b

+

c

)

(

a

2

+

b

2

+

c

2

−

a

b

−

b

c

−

c

a

)

+

3

a

b

c

=

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⇒

(

a

+

b

+

c

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(

a

2

+

b

2

+

c

2

−

(

a

b

+

b

c

+

c

a

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a

b

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⇒

(

1

×

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2

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(

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a

b

c

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=

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⇒

(

2

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)

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3

a

b

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⇒

3

a

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c

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−

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=

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⇒

a

b

c

=

1

6

Now

(

a

2

b

2

+

b

2

c

2

+

c

2

a

2

)

=

(

a

b

+

b

c

+

c

a

)

2

−

2

a

b

2

c

−

2

b

c

2

a

−

2

c

a

2

b

=

(

a

b

+

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c

+

c

a

)

2

−

2

a

b

c

(

b

+

c

+

a

)

=

(

−

1

2

)

2

−

2

×

1

6

×

1

=

1

4

−

1

3

=

−

1

12

Now

a

4

+

b

4

+

c

4

=

(

a

2

+

b

2

+

c

2

)

2

−

2

(

a

2

b

2

+

b

2

c

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+

c

2

a

2

)

=

2

2

−

2

×

(

−

1

12

)

=

4

+

1

6

=

4

1

6

Extension

a

5

+

b

5

+

c

5

=

(

a

3

+

b

3

+

c

3

)

(

a

2

+

b

2

+

c

2

)

−

[

a

3

(

b

2

+

c

2

)

+

b

3

(

c

2

+

a

2

)

+

c

3

(

a

2

+

c

2

)

]

=

3

⋅

2

−

[

a

3

(

b

2

+

c

2

)

+

b

3

(

c

2

+

a

2

)

+

c

3

(

a

2

+

b

2

)

]

Now

a

3

(

b

2

+

c

2

)

+

b

3

(

c

2

+

a

2

)

+

c

3

(

a

2

+

b

2

)

=

a

2

b

2

(

a

+

b

)

+

b

2

c

2

(

b

+

c

)

+

c

2

a

2

(

a

+

c

)

=

a

2

b

2

(

1

−

c

)

+

b

2

c

2

(

1

−

a

)

+

c

2

a

2

(

1

−

b

)

=

a

2

b

2

+

b

2

c

2

+

c

2

a

2

−

(

a

2

b

2

c

+

b

2

c

2

a

+

c

2

a

2

b

)

=

−

1

12

−

a

b

c

(

a

b

+

b

c

+

c

a

)

=

−

1

12

−

1

6

⋅

(

−

1

2

)

=

0

So

a

5

+

b

5

+

c

5

=

6

−

0

=

6

Step-by-step explanation:

8 0
2 years ago
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