Ask a New Question
Menu

Sal is trying to determine which cell phone and service plan to buy for his mother. The first phone costs $100 and $55 per month for unlimited usage. The second phone costs $150 and $51 per month for unlimited usage. How many months will it take for the second phone to be less expensive than the first phone?

The inequality that will determine the number of months, x, that are required for the second phone to be less expensive is
.

The solution to the inequality is
.

Sal’s mother would have to keep the second cell phone plan for at least
months in order for it to be less expensive.

7 answers

To determine how many months it will take for the second phone to be less expensive than the first phone, we can set up an inequality.

Let's define:

  • Cost of the first phone: \( C_1 = 100 + 55x \)
  • Cost of the second phone: \( C_2 = 150 + 51x \)

We want to find the number of months \( x \) such that: \[ C_2 < C_1 \]

Substituting the expressions for \( C_1 \) and \( C_2 \), we get: \[ 150 + 51x < 100 + 55x \]

Now, we can solve this inequality for \( x \):

  1. Subtract \( 51x \) from both sides: \[ 150 < 100 + 4x \]

  2. Subtract 100 from both sides: \[ 50 < 4x \]

  3. Divide both sides by 4: \[ 12.5 < x \]

This can also be rewritten as: \[ x > 12.5 \]

The solution to the inequality indicates that Sal's mother would have to keep the second cell phone plan for at least 13 months (since \( x \) needs to be a whole number).

Thus, the completed statements are:

  • The inequality that will determine the number of months, \( x \), that are required for the second phone to be less expensive is \( 150 + 51x < 100 + 55x \).
  • The solution to the inequality is \( x > 12.5 \).
  • Sal’s mother would have to keep the second cell phone plan for at least 13 months in order for it to be less expensive.
wait what is this The inequality that will determine the number of months, \( x \), that are required for the second phone to be less expensive is \( 150 + 51x < 100 + 55x \).??? The image contains three mathematical inequalities involving the variable ‘x’:

(100 + 55x > 150 + 51x)
(100 + 55x < 150 + 51x)
(100x + 55 > 150x + 51)

I understand now that you’re looking for a representation that matches the form of the inequalities you provided. Let's examine the situation and provide the appropriate inequalities based on your original question regarding the two phone plans.

Given:

  1. The cost of the first phone after \( x \) months is: \[ C_1 = 100 + 55x \]

  2. The cost of the second phone after \( x \) months is: \[ C_2 = 150 + 51x \]

To determine when the second phone becomes less expensive than the first phone, we set up the inequality: \[ C_2 < C_1 \] This translates to: \[ 150 + 51x < 100 + 55x \]

Now, rearranging this inequality:

  1. Subtract \( 51x \) from both sides: \[ 150 < 100 + 4x \]

  2. Subtract 100 from both sides: \[ 50 < 4x \]

  3. Divide both sides by 4: \[ 12.5 < x \] or written in a more common form: \[ x > 12.5 \]

This means that if we want to find a point where the first plan is more expensive than the second one, we would write: \[ 100 + 55x > 150 + 51x \]

This is equivalent to the inequality \( 100 + 55x > 150 + 51x \) in your list.

So:

  • The inequality that will determine the number of months, \( x \), that are required for the second phone to be less expensive is: \[ 100 + 55x > 150 + 51x \]

  • The solution to the inequality is: \[ x > 12.5 \]

  • Sal’s mother would have to keep the second cell phone plan for at least: \[ 13 \text{ months} \]

You can also express the alternative inequality: \[ 100 + 55x < 150 + 51x \] to show when the first plan is less expensive.

If you have any other questions or need further clarification, feel free to ask!

its not 100 + 55x < 150 + 51x

Apologies for any confusion in my previous responses. Let's clarify the inequalities correctly according to your task.

We need to compare the total costs of the two cell phone plans over time.

  1. Cost of the first phone after \( x \) months: \[ C_1 = 100 + 55x \]

  2. Cost of the second phone after \( x \) months: \[ C_2 = 150 + 51x \]

Finding when the second phone is less expensive

We want to set up the inequality so that the cost of the second phone is less than the cost of the first phone: \[ C_2 < C_1 \] This translates to: \[ 150 + 51x < 100 + 55x \]

Rearranging the inequality

To find \( x \):

  1. Subtract \( 51x \) from both sides: \[ 150 < 100 + 4x \]

  2. Subtract 100 from both sides: \[ 50 < 4x \]

  3. Divide both sides by 4: \[ 12.5 < x \] or equivalently, \[ x > 12.5 \]

Answering your original question

  1. The inequality that will determine the number of months, \( x \), that are required for the second phone to be less expensive is: \[ 150 + 51x < 100 + 55x \]

  2. The solution to the inequality is: \[ x > 12.5 \]

  3. Sal’s mother would have to keep the second cell phone plan for at least: \[ 13 \text{ months} \]

Now, if you wanted to consider it in terms of the other inequalities you provided (like \( 100 + 55x > 150 + 51x \)), they are just rearrangements or different ways to express the same situation (showing that the first plan is more expensive).

Thank you for your patience, and I'm here if you have any more questions!

Yuto left his house at 10 a.m. to go for a bike ride. By the time Yuto’s sister Riko left their house, Yuto was already 5.25 miles along the path they both took. If Yuto’s average speed was 0.25 miles per minute and Riko’s average speed was 0.35 miles per minute, over what time period in minutes, t, starting from when Riko left the house, will Riko be behind her brother?

Riko will be behind Yuto when 0 ≤ t <_________
minutes.

To find out how long Riko will be behind Yuto after she leaves the house, we can set up a comparison of their positions over time.

Let's define:

  • \( t \): the time in minutes since Riko left the house.
  • \( D_Y(t) \): the distance Yuto has traveled after \( t \) minutes since Riko left.
  • \( D_R(t) \): the distance Riko has traveled after \( t \) minutes.

Yuto's Distance

Yuto left at 10 a.m. and had already traveled 5.25 miles when Riko left. Since Yuto's speed is 0.25 miles per minute, in \( t \) minutes, he travels an additional: \[ D_Y(t) = 5.25 + 0.25t \]

Riko's Distance

Riko leaves the house and travels at a speed of 0.35 miles per minute. Thus, her distance after \( t \) minutes is: \[ D_R(t) = 0.35t \]

When Will Riko Be Behind Yuto?

Riko will be behind Yuto when: \[ D_R(t) < D_Y(t) \]

Substituting the equations we established: \[ 0.35t < 5.25 + 0.25t \]

Now, we can solve this inequality:

  1. Subtract \( 0.25t \) from both sides: \[ 0.10t < 5.25 \]

  2. Divide both sides by 0.10: \[ t < \frac{5.25}{0.10} = 52.5 \]

Conclusion

Riko will be behind Yuto for the time period when: \[ 0 \leq t < 52.5 \] This means that Riko will be behind Yuto for up to 52 minutes and 30 seconds after she leaves the house.

Riko will be behind Yuto when \( 0 \leq t < 52.5 \) minutes.