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gregori [183]
3 years ago
15

An online shopping club has 4,800 members when it charges $9 per month for membership. For each $1 monthly

Mathematics
1 answer:
Marina CMI [18]3 years ago
8 0

Answer:

The function to represent the revenue received  by the club is

Step-by-step explanation:

Given:

Total members in the Online shopping Club = 13,500

Membership Charge per month =  $8

Increase in the Membership  charge per month = $1

For each $1 monthly increase in membership fee, the club loses approximately  = 500 members

To Find :

The function to represent the revenue received  by the club when x is the price increase

Solution:

For Each month $1 increased is every month which leads to decrease in the members by 500

For 1 month , one dollar  the decrease  is  

for 2 month , 2 dollar increase will lead to decrease of  = 1000  members

So for x months

The decrease in the members  = = 500x

Also the increase in the membership fee will be 8 + x

Now

MARK BRAINLIEST! THANKS HOPE THIS HELPED

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Julli [10]

(a). The solutions are 0 and ⁸/₃.

(b). The solutions are 1 and ¹³/₃.

(c). The equation has no solution.

(d). The only solution is ²¹/₂₀.

(e). The equation has no solution.

<h3>Further explanation</h3>

These are the problems with the absolute value of a function.

For all real numbers x,

\boxed{ \ |f(x)|=\left \{ {{f(x), for \ f(x) \geq 0} \atop {-f(x), for \ f(x) < 0}} \right. \ }

<u>Problem (a)</u>

|4 – 3x| = |-4|

|4 – 3x| = 4

<u>Case 1</u>

\boxed{ \ 4 - 3x \geq 0 \ } \rightarrow \boxed{ \ 4\geq 3x \ } \rightarrow \boxed{ \ x\leq \frac{4}{3} \ }

For 4 – 3x = 4

Subtract both sides by four.

-3x = 0

Divide both sides by -3.

x = 0

Since \boxed{ \ 0\leq \frac{4}{3} \ }, x = 0 is a solution.

<u>Case 2</u>

\boxed{ \ 4 - 3x < 0 \ } \rightarrow \boxed{ \ 4 < 3x \ } \rightarrow \boxed{ \ x > \frac{4}{3} \ }

For -(4 – 3x) = 4

-4 + 3x = 4

Add both sides by four.

3x = 8

Divide both sides by three.

x = \frac{8}{3}

Since \boxed{ \ \frac{8}{3} > \frac{4}{3} \ }, \boxed{ \ x = \frac{8}{3} \ } is a solution.

Hence, the solutions are \boxed{ \ 0 \ and \ \frac{8}{3} \ }  

————————

<u>Problem (b)</u>

2|3x - 8| = 10

Divide both sides by two.

|3x - 8| = 5  

<u>Case 1</u>

\boxed{ \ 3x - 8 \geq 0 \ } \rightarrow \boxed{ \ 3x\geq 8 \ } \rightarrow \boxed{ \ x\geq \frac{8}{3} \ }

For 3x - 8 = 5

Add both sides by eight.

3x = 13

Divide both sides by three.

x = \frac{13}{3}

Since \boxed{ \ \frac{13}{3} \geq \frac{4}{3} \ }, \boxed{ \ x = \frac{13}{3} \ } is a solution.

<u>Case 2</u>

\boxed{ \ 3x - 8 < 0 \ } \rightarrow \boxed{ \ 3x < 8 \ } \rightarrow \boxed{ \ x < \frac{8}{3} \ }

For -(3x – 8) = 5

-3x + 8 = 5

Subtract both sides by eight.

-3x = -3

Divide both sides by -3.

x = 1  

Since \boxed{ \ 1 < \frac{8}{3} \ }, \boxed{ \ x = 1 \ } is a solution.

Hence, the solutions are \boxed{ \ 1 \ and \ \frac{13}{3} \ }  

————————

<u>Problem (c)</u>

2x + |3x - 8| = -4

Subtracting both sides by 2x.

|3x - 8| = -2x – 4

<u>Case 1</u>

\boxed{ \ 3x - 8 \geq 0 \ } \rightarrow \boxed{ \ 3x\geq 8 \ } \rightarrow \boxed{ \ x\geq \frac{8}{3} \ }

For 3x – 8 = -2x – 4

3x + 2x = 8 – 4

5x = 4

x = \frac{4}{5}

Since \boxed{ \ \frac{4}{5} \ngeq \frac{8}{3} \ }, \boxed{ \ x = \frac{4}{5} \ } is not a solution.

<u>Case 2</u>

\boxed{ \ 3x - 8 < 0 \ } \rightarrow \boxed{ \ 3x < 8 \ } \rightarrow \boxed{ \ x < \frac{8}{3} \ }

For -(3x - 8) = -2x – 4

-3x + 8 = -2x – 4

2x – 3x = -8 – 4

-x = -12

x = 12

Since \boxed{ \ 12 \nless \frac{8}{3} \ }, \boxed{ \ x = 12 \ } is not a solution.

Hence, the equation has no solution.

————————

<u>Problem (d)</u>

5|2x - 3| = 2|3 - 5x|  

Let’s take the square of both sides. Then,

[5(2x - 3)]² = [2(3 - 5x)]²

(10x – 15)² = (6 – 10x)²

(10x - 15)² - (6 - 10x)² = 0

According to this formula \boxed{ \ a^2 - b^2 = (a + b)(a - b) \ }

[(10x - 15) + (6 - 10x)][(10x - 15) - (6 - 10x)]] = 0

(-9)(20x - 21) = 0

Dividing both sides by -9.

20x - 21 = 0

20x = 21

x = \frac{21}{20}

The only solution is \boxed{ \ \frac{21}{20} \ }

————————

<u>Problem (e)</u>

2x + |8 - 3x| = |x - 4|

We need to separate into four cases since we don’t know whether 8 – 3x and x – 4 are positive or negative.  We cannot square both sides because there is a function of 2x.

<u>Case 1</u>

  • 8 – 3x is positive  (or 8 - 3x > 0)
  • x – 4 is positive  (or x - 4 > 0)

2x + 8 – 3x = x – 4

8 – x = x – 4

-2x = -12

x = 6

Substitute x = 6 into 8 – 3x ⇒ 8 – 3(6) < 0, it doesn’t work, even though when we substitute x = 6 into x - 4 it does work.

<u>Case 2</u>

  • 8 – 3x is positive  (or 8 - 3x > 0)
  • x – 4 is negative  (or x - 4 < 0)

2x + 8 – 3x = -(x – 4)

8 – x = -x + 4

x – x =  = 4 - 8

It cannot be determined.

<u>Case 3</u>

  • 8 – 3x is negative (or 8  - 3x < 0)
  • x – 4 is positive. (or x - 4 > 0)

2x + (-(8 – 3x)) = x – 4

2x – 8 + 3x = x - 4

5x – x = 8 – 4

4x = 4

x = 1

Substitute x = 1 into 8 - 3x, \boxed{ \ 8 - 3(1) \nless 0 \ }, it doesn’t work. Likewise, when we substitute x = 1 into x – 4, \boxed{ \ 1 - 4 \not> 0 \ }

<u>Case 4</u>

  • 8 – 3x is negative (or 8 - 3x < 0)
  • x – 4 is negative (or x - 4 < 0)

2x + (-(8 – 3x)) = -(x – 4)

2x – 8 + 3x = -x + 4

5x + x = 8 – 4

6x = 4

\boxed{ \ x=\frac{4}{6} \rightarrow x = \frac{2}{3} \ }

Substitute x = \frac{2}{3} \ into \ 8-3x, \boxed{ \ 8 - 3 \bigg(\frac{2}{3}\bigg) \not< 0 \ }, it doesn’t work. Even though when we substitute x = \frac{2}{3} \ into \ x-4, \boxed{ \ \bigg(\frac{2}{3}\bigg) - 4 < 0 \ } it does work.

Hence, the equation has no solution.

<h3>Learn more</h3>
  1. The inverse of a function brainly.com/question/3225044
  2. The piecewise-defined functions brainly.com/question/9590016
  3. The composite function brainly.com/question/1691598

Keywords: hitunglah nilai x, the equation, absolute  value of the function, has no solution, case, the only solution

5 0
3 years ago
Read 2 more answers
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