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ozzi
3 months ago
13

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 100 + 55x > 150 + 51x100 + 55x < 150 + 51x100x+ 55 > 150x+ 51100x+ 55 < 150x+ 51.

The solution to the inequality is x > 2.4x < 2.4x < 12.5x > 12.5.

Sal’s mother would have to keep the second cell phone plan for at least 231213 months in order for it to be less expensive​
Mathematics
2 answers:
Leona [12.6K]3 months ago
8 0

Answer:

Part 1:

The initial phone has a cost of $100 and a monthly charge of $55 for unlimited usage.

Let f(x) denote the cost of the first phone, with x representing the months.

Equation forms:

f(x)=55x+100

The second phone costs $150, with a monthly rate of $51 for unlimited service.

Let g(x) denote the cost of the second phone, with x as the number of months.

Equation forms:

g(x)=51x+150

We need to establish the inequality that shows the months, x, required for the second phone's cost to be lesser. This is given by:

g(x)

51x+150

Part 2:

The inequality solution is:  

51x+150

=> 51x-55x

=> -4x

=> -x

=> x>12.5

Or rounded to 13 months.

Part 3:

For the second phone plan to be less costly, Sal's mother needs to keep it for no less than 13 months.

AnnZ [12.3K]3 months ago
8 0

Answer:

a) The first inequality is 100 + 55x > 150 + 51x;

b) The final inequality results in x > 12.5

c) Sal's mother will need to use the second phone for at least 13 months.

Step-by-step explanation:

a) Let x represent the number of months.

1. The first phone is priced at $100, with a monthly fee of $55 for unlimited use, leading to a total cost of $(100 + 55x) for x months.

2. The second phone costs $150 with a monthly fee of $51 for unlimited use, resulting in a total of $(150 + 51x) for x months.

3. For the second phone to be cheaper, we set up the inequality:

150 + 51x < 100 + 55x

which simplifies to

100 + 55x > 150 + 51x

b) Now solve this:

55x - 51x > 150 - 100

4x > 50

so x > 12.5

c) This means Sal's mother has to retain the second phone for at least 13 months (since x > 12.5).

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If 30,000 cm2 of material is available to make a box with a square base and an open top, what is the largest possible volume (in
Inessa [12570]

Answer:

The highest achievable volume for the box is 2000000 cubic meters.

Step-by-step explanation:

Below is an outline of the volume (V), measured in cubic centimeters, and surface area (A_{s}), measured in square centimeters, for a box featuring a square base:

A_{s} = l^{2}+h\cdot l (1)

V = l^{2}\cdot h (2)

Where:

l - The length of the base's side, in centimeters.

h - The height of the box, in centimeters.

Using (2), we isolate h in the formula:

h = \frac{V}{l^{2}}

Then, we substitute into (1) and simplify the outcome:

A_{s} = l^{2}+ \frac{V}{l}

A_{s}\cdot l = l^{3}+V

V = A_{s}\cdot l -l^{3} (3)

Next, we calculate the first and second derivatives of this expression:

V' = A_{s}-3\cdot l^{2} (4)

V'' = -6\cdot l (5)

If V' = 0 and A_{s} = 30000\,cm^{2}, then we find that the critical value for the base's side length is:

30000-3\cdot l^{2} = 0

3\cdot l^{2} = 30000

l = 100\,cm

Subsequently, we assess this outcome using the second derivative's expression:

V'' = -600

According to Second Derivative Test, this critical value signifies an absolute maximum. Consequently, the largest volume obtainable for the box is:

V = 30000\cdot l - l^{3}

V = 2000000\,cm^{3}

The highest achievable volume for the box is 2000000 cubic meters.

4 0
2 months ago
Identify the range of the function shown in the graph
Leona [12618]

The domain refers to all potential input values, specifically represented by the x-axis on a graph. Conversely, the range includes all possible output values, depicted along the y-axis.


The graph clearly extends horizontally from (-∞,∞) on the x-axis, indicating that its domain is (-∞,∞).


Similarly, it can be seen that the graph stretches vertically from (-∞,∞) on the y-axis, denoting that the range is also (-∞,∞).


This indicates the function includes an infinite array of values. Therefore, there are no limitations on either the domain or the range for this function.

4 0
3 months ago
Read 2 more answers
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