Answer:
Step-by-step explanation:
put x=1 ,y=5
y≥4x
5≥(4×1)
5≥4
which is true.
again 5≤1+2
5≤3
which is not true.
so, no (1,5) is not a solution.
To be a solution it should be true for both.
Answer:
B
Step-by-step explanation:
...
Answer:
130
Step-by-step explanation:
<u>Probability of green:</u>
<u>Number of attempts: </u>
<u>Expected number of landing on green: </u>
- <em>Expected frequency = probability × number of trials</em>
<u>Answer: </u>130 times
F(x) = 5x + 6
let y = f(x)
y = 5x + 6 solving for x.
y - 6 = 5x
5x = y - 6
x = (y - 6)/5
Recall f(x) = y. Implies x = f⁻¹(y)
<span>x = (y - 6)/5
</span>
f⁻¹(y) <span>= (y - 6)/5
</span>
So as well:
f⁻¹(x) <span>= (x - 6)/5</span>
Place a point on (4,1) then count one space up and two spaces to the right and plot the second point there. So (4,1) and (6,2). That should graph the line automatically when you place two points