The solutions are (2.1925, 5.1925) and (-3.1925, -0.1925)
<em><u>Solution:</u></em>
Given that,
<em><u>We have to substitute eqn 1 in eqn 2</u></em>
Substitute x = 2.1925 in eqn 1
y = 2.1925 + 3
y = 5.1925
Substitute x = -3.1925 in eqn 1
y = -3.1925 + 3
y = -0.1925
Thus the solutions are (2.1925, 5.1925) and (-3.1925, -0.1925)
Answer:
Age 42 salary 30,000
Step-by-step explanation:
He had to restart his medical training, therefore, making him have a lower salary regardless of his age, making this point an obvious outlier
90 ft.......hope this helps
Recognize that both 0.96 and 0.144 are divisible by 12:
(0.96/12) / (0.144/12) = 0.08 / 0.012. This reduces to 0.02 / 0.003, or
20/3 or approx. 6.666.
You could also begin by eliminating the decimal fractions. Mult. 0.96 and 0.144 each by 1000 results in 960/144.
Since both 960 and 144 can be divided evenly by 24, we get 40 and 6.
40/6 = 20/3, or approx. 6.666, as before.
The set must have a constant additive rate of change.
Answer: Option A.
<u>Explanation:</u>
A function has an additive rate of change if there is a consistent distinction between any two successive information and yield esteems. From the primary table we can watch a consistent distinction of - 6 among the y-values and a steady contrast of 2 among the x-values.
A linear function has a consistent additive rate of change, while a nonlinear capacity doesn't. For a table of qualities to be direct, the needy variable must have a steady pace of progress as the autonomous variable increments by 1.
