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Ganezh [65]
2 years ago
15

For each of the number lines, write an absolute value equation that has the following solution set. 26 and m

Mathematics
1 answer:
alexira [117]2 years ago
6 0

On a number line, an absolute value equation that has the given solution set is |m - 4| = 2.

<h3>How to write the absolute value equation?</h3>

By critically observing the given question, we can infer and logically deduce that the solution sets for this absolute value equation is given by:

m = {2, 6}

Next, we would calculate the mean of the solution sets as follows:

m₁ = (2 + 6)/2

m₁ = 8/2

m₁ = 4.

Also, we would calculate the difference in the solution sets as follows:

m₂ = (6 - 2)/2

m₂ = 4/2

m₂ = 2.

Mathematically, the absolute value equation is given by:

|m - m₁| - m₂ = 0

|m - 4| - 2 = 0

|m - 4| = 2.

Read more on absolute value equation here: brainly.com/question/27197258

#SPJ1

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daser333 [38]

Answer:

a) Number of projects in the first year = 90

b) Earnings in the twelfth year = $116500

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Step-by-step explanation:

Given that:

Number of projects done in fourth year = 129

Number of projects done in tenth year = 207

There is a fixed increase every year.

a) To find:

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This problem is nothing but a case of arithmetic progression.

Let the first term i.e. number of projects done in first year = a

Given that:

a_4=129\\a_{10}=207

Formula for n^{th} term of an Arithmetic Progression is given as:

a_n=a+(n-1)d

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a_4=129=a+(4-1)d \\\Rightarrow 129=a+3d .....(1)\\a_{10}=207=a+(10-1)d \\\Rightarrow 207=a+9d .....(2)

Subtracting (2) from (1):

78 = 6d\\\Rightarrow d =13

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129 =a+3\times 13\\\Rightarrow a =129-39\\\Rightarrow a =90

<em>Number of projects in the first year = 90</em>

<em></em>

<em>b) </em>

Number of projects in the twelfth year =

a_{12} = a+11d\\\Rightarrow a_{12} = 90+11\times 13 =233

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Earnings in the twelfth year = 233 \times 500 = $116500

Sum of an AP is given as:

S_n=\dfrac{n}{2}(2a+(n-1)d)\\\Rightarrow S_{12}=\dfrac{12}{2}(2\times 90+(12-1)\times 13)\\\Rightarrow S_{12}=6\times 323\\\Rightarrow S_{12}=1938

It gives us the total number of projects done in 12 years = 1938

Total money earned in 12 years = 500 \times 1938 = $969000

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Answer:

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