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Tatiana [17]
1 year ago
11

Which number line represents the solution set for the inequality 3(8 – 4x) < 6(x – 5)?

Mathematics
1 answer:
kogti [31]1 year ago
5 0

The solution will be a number line from negative 5 to 5 in increments of 1. An open circle is at 3 and a bold line starts at 3 and is pointing to the right.

<h3>Inequality expression</h3>

Inequalities are equations that are separated by an equal sign. Given the inequality expression

3(8 – 4x) < 6(x – 5)

Expand

3(8) - 3(4x) < 6x - 6(5)

24 - 12x < 6x - 30

Collect the like terms

-12x - 6x < -30 - 24

-18x < -54

x > 3

Hence the solution will be a number line from negative 5 to 5 in increments of 1. An open circle is at 3 and a bold line starts at 3 and is pointing to the right.

Learn more on inequality here: brainly.com/question/11613554

#SPJ1

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Answer:

Let v(t) be the velocity of the car t hours after 2:00 PM. Then \frac{v(1/3)-v(0)}{1/3-0}=\frac{50 \:{\frac{mi}{h} }-30\:{\frac{mi}{h} }}{1/3\:h-0\:h} = 60 \:{\frac{mi}{h^2} }.  By the Mean Value Theorem, there is a number c such that 0 < c with v'(c)=60 \:{\frac{mi}{h^2}}. Since v'(t) is the acceleration at time t, the acceleration c hours after 2:00 PM is exactly 60 \:{\frac{mi}{h^2}}.

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The Mean Value Theorem says,

Let be a function that satisfies the following hypotheses:

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Then there is a number c in (a, b) such that

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Note that the Mean Value Theorem doesn’t tell us what c is. It only tells us that there is at least one number c that will satisfy the conclusion of the theorem.

By assumption, the car’s speed is continuous and differentiable everywhere. This means we can apply the Mean Value Theorem.

Let v(t) be the velocity of the car t hours after 2:00 PM. Then v(0 \:h) = 30 \:{\frac{mi}{h} } and v( \frac{1}{3} \:h) = 50 \:{\frac{mi}{h} } (note that 20 minutes is 20/60=1/3 of an hour), so the average rate of change of v on the interval [0 \:h, \frac{1}{3} \:h] is

\frac{v(1/3)-v(0)}{1/3-0}=\frac{50 \:{\frac{mi}{h} }-30\:{\frac{mi}{h} }}{1/3\:h-0\:h} = 60 \:{\frac{mi}{h^2} }

We know that acceleration is the derivative of speed. So, by the Mean Value Theorem, there is a time c in (0 \:h, \frac{1}{3} \:h) at which v'(c)=60 \:{\frac{mi}{h^2}}.

c is a time time between 2:00 and 2:20 at which the acceleration is 60 \:{\frac{mi}{h^2}}.

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