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Sophia College Algebra Milestone 2- Actual Questions with Answers 2024/2025. $15.49   Add to cart

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Sophia College Algebra Milestone 2- Actual Questions with Answers 2024/2025.

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  • SOPHIA ALGEBRA
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  • SOPHIA ALGEBRA

Sophia College Algebra Milestone 2- Actual Questions with Answers 2024/2025.

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  • September 7, 2024
  • 13
  • 2024/2025
  • Exam (elaborations)
  • Questions & answers
  • SOPHIA ALGEBRA
  • SOPHIA ALGEBRA
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BrieCouture
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1
Consider the following subset of the real number line




UNIT 2 — MILESTONE 2
How can this set be expressed using inequalities?



-3 ≤ x < 1



-3 ≥ x > 1



-3 < x ≤ 1



-3 > x ≥ 1


RATIONALE




CONCEPT
Introduction to Inequalities 2

What is the solution set for the following inequality?

2x + 8x ≥ 15 + 14x + 9



x ≤ -6



x ≤ -0.85

, x ≥ -6 x



≥ -0.85



RATIONALE
Before solving for x, make sure that each side of the inequality is fully simplified. On the right side, we will add 15 and 9.

On the right side, 15 plus 9 is 24. On the left side, we can combine 2x and 8x.

2x plus 8x is 10x. Now we can begin to solve for x by applying inverse operations to both sides of the inequality. First, subtract 14x from both
sides.

Subtracting 14x from both sides cancels the 14x term on the right side of the inequality, leaving x terms on only the left side. Finally, divide both
sides of the inequality by -4. Remember that the sign of the inequality sign changes whenever you multiply or divide by a negative number!
The solution to the inequality is x ≤ -6.


CONCEPT
Solve Linear Inequalities 3

John is driving at a constant speed of 40 miles per hour.

How long does it take him to travel 50 miles?




An hour and 48 minutes

An hour and 15 minutes




An hour and 8 minutes

An hour and 30 minutes


RATIONALE
To find how long it will take John to travel 50 miles, we can use the distance, rate, time formula and solve for time. Plug in 50
miles for the distance, and 40 miles per hour for the rate, or speed.

Once we have plugged in the values, we need to write miles per hour as a fraction: 40 miles over 1 hour.


When dividing by a fraction, we can change this into a multiplication problem and multiply by the reciprocal of 40 miles per 1
hour, which would be 1 hour over 40 miles.


Rewrite 50 miles as a fraction over 1, and multiply this by the reciprocal of 40 mph. Next, multiply the numerators and
denominators of the fractions.

Multiplying across the numerator and denominator, produces 50 over 40. The units of miles cancel, so we are left with hours.
Finally, divide 50 by 40.
It will take John 1.25 hours. However, because we must express our answer in hours and minutes, we must convert 0.25 hours
to minutes.

Using the conversion factor, 60 minutes to 1 hour, evaluate the fractions by multiplying the numerators and denominators.

Multiplying across the numerator and denominator produces 0.25 times 60, or 15. Units of hours cancel, leaving only minutes.
It will take John 1 hour and 15 minutes to travel 50 miles at a constant speed of 40 miles per hour.

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