Question 1
Part A:
The water balloon's height is increasing between the ordered pairs (0, 26) and (2, 34). Therefore, the height of the water balloon is increasing during the interval \(0 \leq x < 2\).
Part B:
The water balloon's height stays the same between the ordered pairs (2, 34) and (4, 34). Hence, the height of the water balloon is constant during the interval \(2 \leq x < 4\).
Part C:
The height of the water balloon is decreasing the fastest between the ordered pairs (10, 0) and (8, 30). The change from 30 feet at 8 seconds to 0 feet at 10 seconds occurs over 2 seconds, resulting in a steep decline. Thus, the interval with the fastest decrease is from \(8\) seconds to \(10\) seconds, as the height drops from \(30\) feet to \(0\) feet, representing a significant drop in a short time.
Part D:
Based on the pattern from the given data, after 12 seconds, the height of the water balloon is predicted to remain at \(0\) feet. The balloon will have already hit the ground at 10 seconds, and remains at \(0\) feet beyond that time for any time intervals greater than or equal to 10 seconds.
Question 2
Part A:
To find the test average for the math class after completing test 2, we substitute \(x = 2\) into the function:
\[ f(2) = 0.2(2) + 79 = 0.4 + 79 = 79.4 \]
Thus, the average test score in math after test 2 is 79.4.
Part B:
To determine the test average for the science class after completing test 2, we refer to the table where \(g(2) = 84\). Therefore, the average test score in science after test 2 is 84.
Part C:
To find the average score after completing test 4 for both math and science, we first calculate \(f(4)\):
\[ f(4) = 0.2(4) + 79 = 0.8 + 79 = 79.8 \]
For science, we need to identify \(g(4)\). The pattern shows that the average test score decreases by 2 points per test:
- \(g(1) = 86\)
- \(g(2) = 84\)
- \(g(3) = 82\)
- Therefore, \(g(4) = 80\).
Now, we compare the two scores after test 4:
- Math average = 79.8
- Science average = 80
Hence, the science class had a higher average after completing test 4, with an average score of 80 compared to the math class's score of 79.8.
Question 3
Part A:
To find the slope of the function, we can take two points from the table:
- Point 1: (0, 325)
- Point 2: (5, 400)
Using the slope formula \((y_2 - y_1)/(x_2 - x_1)\):
\[ \text{slope} = \frac{400 - 325}{5 - 0} = \frac{75}{5} = 15 \]
The slope of 15 means that the account balance increases by $15 for each day that passes.
Part B:
Using point-slope form, we use the point (0, 325):
-
Point-Slope form: \(y - y_1 = m(x - x_1)\): \[ g(x) - 325 = 15(x - 0) \implies g(x) = 15x + 325 \]
-
Slope-Intercept form: \(g(x) = 15x + 325\).
-
Standard form: \(15x - g(x) + 325 = 0\) (or, \(15x - y + 325 = 0\)).
Part C:
The equation using function notation is: \[ g(x) = 15x + 325 \]
Part D:
To find the balance in the bank account after 12 days, we substitute \(x = 12\) into the equation:
\[ g(12) = 15(12) + 325 = 180 + 325 = 505 \]
The balance in the bank account after 12 days is $505.
Question 4
Part A:
The function \(g(x) = |x - 7|\) is an absolute value function with a vertex at the point (7, 0). The domain of the function is all real numbers, expressed as \((-∞, ∞)\). The range is all non-negative real numbers, expressed as \([0, ∞)\).
Part B:
The transformation from \(f(x) = |x|\) to \(h(x) = |x| + 2\) indicates a vertical shift upwards by \(2\) units. The vertex of \(h(x)\) moves from (0, 0) for \(f(x)\) to (0, 2) for \(h(x)\). Consequently, the range of \(h(x)\) becomes \([2, ∞)\) instead of \([0, ∞)\).