Part A
Question 1
Let’s say Sheila decides to buy shirts. The equation she would use to solve for \( x \) (the number of shirts) with her $150 gift card is:
\[ 18.75x = 150 \]
Question 2
To determine if \( x = 9 \) is a possible solution to this equation, Sheila can substitute \( x \) into the equation:
\[ 18.75(9) = 168.75 \]
Since $168.75 exceeds $150, \( x = 9 \) is not a viable solution.
Question 3
The steps for Sheila to solve the equation \( 18.75x = 150 \) are:
- Divide both sides of the equation by 18.75: \[ x = \frac{150}{18.75} \]
- Calculate \( x \): \[ x = 8 \]
This tells her she can buy 8 shirts.
Question 4
To verify if her solution of \( x = 8 \) is correct, Sheila can check:
\[ 18.75 \times 8 = 150 \]
Since the result is equal to the total amount on her gift card, the solution is correct.
Part B
Question 5
Let’s pick a scarf and a hat. The inequality would be:
\[ (12.50 + 25.00) \cdot x \leq 150 \] or simplified: \[ 37.50x \leq 150 \]
Question 6
To determine if \( x = 1.5 \) is a viable solution, Sheila should substitute \( x \) into the inequality:
\[ 37.50(1.5) = 56.25 \leq 150 \]
Since this statement is true, \( x = 1.5 \) is a viable solution.
Question 7
To find the solution set for the inequality \( 37.50x \leq 150 \), divide both sides by 37.50:
\[ x \leq \frac{150}{37.50} = 4 \]
Thus the solution set is \( x \in [0, 4] \).
Question 8
On a number line from -5 to 5, the shaded region representing \( x \in [0, 4] \) will include points 0, 1, 2, 3, and 4.
Part C
Question 9
For the equations \( 5x + 13(x+4) - 211(x - 1) = 11 \), \( 5(x + 1) = 4x + 3 \), and \( 2(x + 2) \):
To solve these equations, it is crucial to isolate variable \( x \) on one side. However, verifying which yield the same solution requires careful algebraic manipulation.
Question 10
An example of a true equation can be: \( 3 + 2 = 5 \) (true). An example of a false inequality: \( 5 < 2 \) (false). The equation is true because the left side equals the right side, while the inequality is false since 5 is not less than 2.
Question 11
Choosing equations: \( 4x + 6 = 12 \), \( 3x < 9 \), and \( 2x + 4 \geq 10 \):
- The conjunction results in specific values of \( x \) satisfying all three.
- The disjunction allows for a wider range where any of the statements holds true.
Comparing both sets shows that “and” leads to stricter criteria, while “or” encompasses broader possibilities.
Question 12
Rewriting \( 9x - 3 = x^2 \) as two equations:
- \( 9x - 3 = 0 \)
- \( x^2 = 0 \)
The significance of \( x^2 = 0 \) indicates that the solutions are those where \( x = 0 \).
Question 13
Rearranging \( F = 9.5C + 32 \) yields \( C = \frac{F - 32}{9.5} \). The first formula is useful when converting Celsius to Fahrenheit, while the second does the reverse.
Question 14
Part A
The equation for the temperature drop per hour is:
\[
\text{Rate} = \frac{\text{Initial temp} - \text{Final temp}}{\text{Time interval}} = \frac{82 - 62}{5} = 4°F \text{ per hour}
\]
Part B
Continuing to drop at the same rate, the temperature at 9 p.m. would be:
\[
62 - 4(2) = 54°F
\]
Question 15
Using \( 35.75x + 10 = 546.25 \):
- First, subtract 10 from both sides: \[ 35.75x = 536.25 \]
- Then divide by 35.75: \[ x = \frac{536.25}{35.75} = 15 \]
So, Pria tutored Jake for 15 hours, as this calculation confirms.
If you have additional questions or need more detailed explanations, feel free to ask!
1 answer