Do you understand what I did here?
<em>f(d)=86,400·d</em>
if you set 1 day (d=1) you get f(1)=86,400 sec
if you set 1 day (d=2) you get f(2)=172,800 sec
...etc.
Answer:
500
Step-by-step explanation:
Rate at which first ticket printer works is 450/9 = 50 tickets per minute
Speed of second ticket printer is twice than that of the first.
So, speed of second printer is 100 tickets per minute.
In 5 minutes,second printer made 500 tickets
10 times 20,000 equals 200,000. So the answer is C, the value of the 2 is 10 times greater