Question

A movie theater offers two options for membership. The first option costs $10 per ticket but an initial fee of $15. The second option costs $7.50 per ticket but an initial fee of $45 . After how many tickets are both options the same price?

Answer 1

Two pumps are being filled by two different pumps. The first vat has 161616 gallons is being filled at a rate of 19 gallons per second. The second vat has 37 gallons and is being filled at a rate of 15 gallons per second. After how many seconds will both vats have the same amount of liquid?

Answer 2

Two pumps are being filled by two different pumps. The first vat has 16 gallons is being filled at a rate of 19 gallons per second. The second vat has 37 gallons and is being filled at a rate of 15 gallons per second. After how many seconds will both vats have the same amount of liquid?
(i had meant 16 not 161616 srry)

Answer 3

There are two different vats. The first vat has 20 gallons is being filled at a rate of 3 gallons per minute. The second vat has 104 gallons and is leaking fluid at a rate of 2 gallons per minute. After how many seconds will both vats have the same amount of liquid?

Answer 4

1008 seconds as minutes...? 😭

Answer 5

A movie theater offers a special pricing for membership. The normal cost is $11 per ticket. The membership costs $6 per ticket but an initial fee of $25 . After how many tickets are both options the same price?

Answer 6

To find out after how many tickets both membership options cost the same, we can set up an equation based on the total cost for each option.

Let \( x \) represent the number of tickets purchased.

For the first membership option:
- Initial fee: \( 15 \)
- Cost per ticket: \( 10 \)
- Total cost for \( x \) tickets:
\[
C_1 = 15 + 10x
\]

For the second membership option:
- Initial fee: \( 45 \)
- Cost per ticket: \( 7.50 \)
- Total cost for \( x \) tickets:
\[
C_2 = 45 + 7.50x
\]

We want to find the value of \( x \) where both costs are equal:
\[
15 + 10x = 45 + 7.50x
\]

Now, we'll solve this equation for \( x \).

1. Subtract \( 7.50x \) from both sides:
\[
15 + 10x - 7.50x = 45
\]
\[
15 + 2.5x = 45
\]

2. Subtract \( 15 \) from both sides:
\[
2.5x = 30
\]

3. Divide both sides by \( 2.5 \):
\[
x = \frac{30}{2.5} = 12
\]

Thus, the two membership options will cost the same after purchasing **12 tickets**.

Answer 7

Let \( t \) represent the time in seconds.

For the first vat:
- Initial amount: \( 161616 \) gallons
- Filling rate: \( 19 \) gallons per second

The amount of liquid in the first vat after \( t \) seconds is:
\[
A_1 = 161616 + 19t
\]

For the second vat:
- Initial amount: \( 37 \) gallons
- Filling rate: \( 15 \) gallons per second

The amount of liquid in the second vat after \( t \) seconds is:
\[
A_2 = 37 + 15t
\]

We want to find when both vats have the same amount of liquid:
\[
161616 + 19t = 37 + 15t
\]

Now, we'll solve the equation for \( t \).

1. Subtract \( 15t \) from both sides:
\[
161616 + 19t - 15t = 37
\]
\[
161616 + 4t = 37
\]

2. Subtract \( 161616 \) from both sides:
\[
4t = 37 - 161616
\]
\[
4t = -161579
\]

3. Divide both sides by \( 4 \):
\[
t = \frac{-161579}{4} = -40394.75
\]

Since this result is negative, it indicates that the second vat will never catch up to the first vat. The first vat is initially much larger and is filling at a higher rate, so the second vat will never equal the volume of the first vat.

**Conclusion:** After \( t \) seconds, the two vats will never have the same amount of liquid.

Answer 8

No problem! Let’s solve the problem with the corrected numbers:

Let \( t \) represent the time in seconds.

For the first vat:
- Initial amount: \( 16 \) gallons
- Filling rate: \( 19 \) gallons per second

The amount of liquid in the first vat after \( t \) seconds is:
\[
A_1 = 16 + 19t
\]

For the second vat:
- Initial amount: \( 37 \) gallons
- Filling rate: \( 15 \) gallons per second

The amount of liquid in the second vat after \( t \) seconds is:
\[
A_2 = 37 + 15t
\]

We want to find when both vats have the same amount of liquid:
\[
16 + 19t = 37 + 15t
\]

Now, let's solve the equation for \( t \).

1. Subtract \( 15t \) from both sides:
\[
16 + 19t - 15t = 37
\]
\[
16 + 4t = 37
\]

2. Subtract \( 16 \) from both sides:
\[
4t = 37 - 16
\]
\[
4t = 21
\]

3. Divide both sides by \( 4 \):
\[
t = \frac{21}{4} = 5.25
\]

Thus, after **5.25 seconds**, both vats will have the same amount of liquid.

Answer 9

To find out when both vats will have the same amount of liquid, we can set up equations that represent the amount of liquid in each vat over time.

Let \( t \) be the time in minutes.

For the **first vat**:
- Initial amount: \( 20 \) gallons
- Filling rate: \( 3 \) gallons per minute

The amount of liquid in the first vat after \( t \) minutes is:
\[
A_1 = 20 + 3t
\]

For the **second vat**:
- Initial amount: \( 104 \) gallons
- Leaking rate: \( 2 \) gallons per minute

The amount of liquid in the second vat after \( t \) minutes is:
\[
A_2 = 104 - 2t
\]

We want to find the time \( t \) when both vats have the same amount of liquid:
\[
20 + 3t = 104 - 2t
\]

Now, let's solve this equation for \( t \).

1. Add \( 2t \) to both sides:
\[
20 + 3t + 2t = 104
\]
\[
20 + 5t = 104
\]

2. Subtract \( 20 \) from both sides:
\[
5t = 104 - 20
\]
\[
5t = 84
\]

3. Divide both sides by \( 5 \):
\[
t = \frac{84}{5} = 16.8 \text{ minutes}
\]

To convert this time to seconds, multiply by 60:
\[
t = 16.8 \times 60 = 1008 \text{ seconds}
\]

Thus, after **1008 seconds**, both vats will have the same amount of liquid.

Answer 10

I apologize for the oversight in my response! Let's convert the seconds back to minutes correctly.

The time we calculated is \( 1008 \) seconds. To convert seconds to minutes, we divide by \( 60 \):

\[
t = \frac{1008}{60} = 16.8 \text{ minutes}
\]

Thus, we find:

- \( 1008 \) seconds is equivalent to **16 minutes and 48 seconds** (since \( 0.8 \times 60 = 48 \) seconds).

So, the final answer is that after **16 minutes and 48 seconds**, both vats will have the same amount of liquid.

Thank you for your patience, and I apologize for the confusion.

Answer 11

Let \( x \) represent the number of tickets purchased.

For the normal ticket price:
- Cost per ticket: \( 11 \)
- Total cost for \( x \) tickets:
\[
C_1 = 11x
\]

For the membership option:
- Initial fee: \( 25 \)
- Cost per ticket: \( 6 \)
- Total cost for \( x \) tickets:
\[
C_2 = 25 + 6x
\]

We want to find the value of \( x \) where both costs are equal:
\[
11x = 25 + 6x
\]

Now, let's solve the equation for \( x \).

1. Subtract \( 6x \) from both sides:
\[
11x - 6x = 25
\]
\[
5x = 25
\]

2. Divide both sides by \( 5 \):
\[
x = \frac{25}{5} = 5
\]

Thus, the two pricing options will cost the same after purchasing **5 tickets**.