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iVinArrow [24]
3 years ago
11

Which compound sentence has its solution set shown on the graph below

Mathematics
1 answer:
musickatia [10]3 years ago
8 0

A compound inequality is a sentence with two inequality statements joined either by the word “or” or by the word “and.” “And” indicates that both statements of the compound sentence are true at the same time.

“Or” indicates that, as long as either statement is true, the entire compound sentence is true.

Now as shown in the graph, the solution inequality of the graph is :

x > 3 and x < 5   [please note, circles in the graph indicate exclusion, dots indicate inclusion. in the graph given circles are shown, so it depicts exclusion]

Now let's solve each option to find if it fits in with the above inequality

Option 1 : 2x-4 > 6 or 3x < 9

⇒ x > 5 or x < 3

Option 2 : 2x - 4 < 6 and 3x > 9

⇒ x < 5 and x > 3

Option 3 : 3x + 8 > -7 or -4x < 12

⇒ 3x > -15 or x < -3

⇒ x > -5 or x < -3

Option 4 : 3x + 8 < -7 and -4x > 12

⇒ 3x < -15 and x > -3

⇒ x < -5 and x > -3


So the compound sentence in option 2 : 2x - 4 < 6 and 3x > 9

has its solution set on the graph.


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The capital formation of the investment function over a given period is the

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  • (a) The capital formation from the end of the second year to the end of the fifth year is approximately <u>298.87</u>.

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(a) The given investment function is presented as follows;

I(t) = 100 \cdot e^{0.1 \cdot t}

(a) The capital formation is given as follows;

\displaystyle Capital = \int\limits {100 \cdot e^{0.1 \cdot t}} \, dt =1000 \cdot  e^{0.1 \cdot t}} + C

From the end of the second year to the end of the fifth year, we have;

The end of the second year can be taken as the beginning of the third year.

Therefore,  for the three years; Year 3, year 4, and year 5, we have;

\displaystyle Capital = \int\limits^5_3 {100 \cdot e^{0.1 \cdot t}} \, dt \approx 298.87

The capital formation from the end of the second year to the end of the fifth year, C ≈ 298.87

(b) When the capital stock exceeds $100,000, we have;

\displaystyle  \mathbf{\left[1000 \cdot  e^{0.1 \cdot t}} + C \right]^t_0} = 100,000

Which gives;

\displaystyle 1000 \cdot  e^{0.1 \cdot t}} - 1000 = 100,000

\displaystyle \mathbf{1000 \cdot  e^{0.1 \cdot t}}} = 100,000 + 1000 = 101,000

\displaystyle e^{0.1 \cdot t}} = 101

\displaystyle t = \frac{ln(101)}{0.1} \approx 46.15

The number of years before the capital stock exceeds $100,000 ≈ <u>46.15 years</u>.

Learn more investment function here:

brainly.com/question/25300925

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