It’s is 1800
Hope this helps
Have a great day/night
Answer:
12.56 mm²
Step-by-step explanation:

Plug the given values into the area equation

Therefore, the area of the circle is 12.56 mm².
I hope this helps!
You need to use basic algebra for this.
For this I’ll use o as the items and p for the payment. First you need to find out how long it took for all the items to scan, so if it took each item 2 seconds to be scanned you need to times the total number of items (o) by two e.g. o x 2 = 62 items times two seconds which is equivalent to 62 seconds (1.02 minutes) after this step you need to minus the total time it took to scan the items for the transaction time (2 minutes) e.g. 2.00 - 1.02 = 2.58 minutes.
Hope this helped :)
Answer:
43.35
Step-by-step explanation:
Answer:
R = sqrt[(IWL)^2/(E^2 - I^2)] or R = -sqrt[(IWL)^2/(E^2 - I^2)]
Step-by-step explanation:
Squaring both sides of equation:
I^2 = (ER)^2/(R^2 + (WL)^2)
<=>(ER)^2 = (I^2)*(R^2 + (WL)^2)
<=>(ER)^2 - (IR)^2 = (IWL)^2
<=> R^2(E^2 - I^2) = (IWL)^2
<=> R^2 = (IWL)^2/(E^2 - I^2)
<=> R = sqrt[(IWL)^2/(E^2 - I^2)] or R = -sqrt[(IWL)^2/(E^2 - I^2)]
Hope this helps!