<span>A.Two acute angles and two obtuse angles
Hope this helps!
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Since (f/g)(x) = f(x)/g(x) for x to be in the domain of (f/g)(x) it must be in the domain of f and in the domain of g. You also need to insure that g(x) is not zero since f(x) is divided by g(x). Thus there are 3 conditions. x must be in the domain of f: f(x) = 3x -5 and all real numbers x are in the domain of x.
Given f(x) = 2x + 3 and g(x) = –x2 + 5, find ( f o f )(x).
( f o f )(x) = f ( f (x))
= f (2x + 3)
= 2( ) + 3 ... setting up to insert the input
= 2(2x + 3) + 3
= 4x + 6 + 3
= 4x + 9
Given f(x) = 2x + 3 and g(x) = –x2 + 5, find (g o g)(x).
(g o g)(x) = g(g(x))
= –( )2 + 5 ... setting up to insert the input
= –(–x2 + 5)2 + 5
= –(x4 – 10x2 + 25) + 5
= –x4 + 10x2 – 25 + 5
= –x4 + 10x2 – 20
The discount is 15%, multiply the price y 15%:
5000 x 0.15 = 750
The discount is 750
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Thus the correct answer is the last option .
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Answer:
0.99945
Step-by-step explanation:
The probability that a bridge hand of 13 cards would contain at least one card that is ten or higher would be the inverse of the probability that all of the 13 cards would not contain any 10 or higher, in other words all of them are 9 or lower. From 2 to 9 we have 8 cards of 4 suits like that, so a total of 32 out of 52 cards
The probability of this to happen is
In the first draw, chance are 32/52
In the 2nd draw, chance are 31/51
In the 3rd draw, chance are 30/50
...
In the 13th draw, chance are 20/40
So the total probability is
So the probability that a randomly selected bridge hand is a Yarborough is
1 - 0.000547 = 0.99945