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2 years ago

Find the range of values which satisfy the following:

Mathematics

1 answer:

egoroff_w [7]2 years ago

6 0

The range of values that satisfy the given inequalities, 8x - 11 > 5 and 15(6x+2) > -150, is x > -2 OR x > 2

<h3>Solving Linear Inequalities: Determining range of values</h3>

From the question, we are to determine the range of values which satisfy the given inequalities.

The given inequalities are

8x - 11 > 5 and 15(6x+2) > -150

To determine the range of values that satisfy the inequalities, we will solve each of the inequalities separately

Solving 8x - 11 > 5

Add 11 to both sides

8x - 11 + 11 > 5 + 11

8x > 16

Divide both sides by 8

8x/8 > 16/8

x > 2

Solving 15(6x+2) > -150

Divide both sides by 15

15(6x+2)/15 > -150/15

6x + 2 > -10

Subtract 2 from both sides

6x + 2 - 2 > -10 - 2

6x > -12

Divide both sides by 6

6x/6 > -12/6

x > -2

Putting the two solutions together,

x > -2 OR x > 2

Hence, the solution is x > -2 OR x > 2

Learn more on Solving inequalities here: brainly.com/question/246993

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Answer:The pool must have been the same depth at the start of the interval as it was at the end of the interval.

Step-by-step explanation:

The average rate of change is calculated as:

[final value - initial value] / time interval.

Then, the average rate of change does not take into account intermediates values, and you cannot draw any conclusion about such intermediate values.

In the given case you have:

average rate of change in depth = [final depth - initial depth] / 2 weeks.

0 = [final depth - initial depth] / 2 weeks.

⇒ 0 = final depth - initial depth

⇒ final depth = initial depth.

That is why the conclusion is the second statement of the answer choices: the pool must have been the same depth at the start of the interval as it was at the end of the interval.

In between the pool might have been deeper, more shallow, empty or change in any form, since the average rate of change does not tell the full history but only the net change.

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Gold-192 is a radioactive isotope with a half-life of 4.95 hours. How long would it take for a 25 gram sample of Gold-192 to dec

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Answer:

Step-by-step explanation:

The formula you will need for this is:

$A=A_0(\frac{1}{2})^{\frac{t}{H}$ where A is the amount after some decay has happened, A₀ is the intial amount, t is the time in hours, and H is the half-life in hours. Those values for us are:

A = 2 g

A₀ = 25 g

H = 4.95 hrs and

t = ?

Filling in:

$2=25(\frac{1}{2})^{\frac{t}{4.95}$ Keep in mind that, because of the nature of the exponential form of this equation, you CANNOT simply multiply the 25 by the 1/2. Exponential equations don't work that way. Begin instead by dividing both sides by 25 to get