Leannie bought 5 of Dr. Mumin’s book and 3 of Dan Moyer’s book
Answer:
The answer is "Options A, B, and E represent mutually exclusive events".
Step-by-step explanation:
Two occurrences that can happen immediately called mutually incompatible. Let's now glance at our options and figure out where the statements are mutually incompatible events.
In Option A: You could see that landing on an unwanted portion and arriving on 2 are events that are locally incompatible, even as undesirable portion contains 3 and 4, and 2 were shaded.
In Option B: Arriving on a shaded part and falling on 3 are also mutually incompatible because there are 3 on a windows azure.
In Option C: A darkened portion and an increasing amount can land while 2 would be an even number as well as on the shaded portion. That number is very much the same.
In Option D: At the same time as 4 is greater than 3 and it is situated upon an undistressed section, landing and attracting a number larger than 3 can happen.
In Option E: Landing on a shaded part and landing on even a shaded part is an excluding event, since shaders may either be shaded or unlit.
I hope this helps you
let's find AB
AB=square root of 2^2+3^2
AB=square root of 13
cosO=2/sqrt13
sinO=3/sqrt13
tgO=3/2
ctgO=2/3
Answer:
see explanation
Step-by-step explanation:
If A +B = 45° then tan(A+B) = tan45° = 1
Expanding (1 + tanA)(1 + tanB)
= 1 + tanA + tanB + tanAtanB → (1)
Using the Addition formula for tan(A + B)
tan(A+B) =
= 1 ← from above
Hence
tanA + tanB = 1 - tanAtanB ( add tanAtanB to both sides )
tanA + tanB + tanAtanB = 1 ( add 1 to both sides )
1 + tanA + tanB + tanAtanB = 2
Then from (1)
(1 + tanA)(1 + tanB) = 2 ⇒ proven
First look to see if the largest denominator is divisible by the smaller denominator. In this case it is. So the lcd is the largest denominator 15.