I think that x^2-9 (difference of 2 squares), 216x^3 +729 (sum of 2 cubes), and the other two are none
Not entirely sure
We can find the number of extra clubs in the deck from the given probability, since the probability of drawing a diamond and a club is

but this would imply that

, which suggests we're taking 4 clubs out of the original deck and contradicts the problem statement that there are "too many clubs".
Perhaps the question is providing the probability of drawing a diamond, THEN a club, or vice versa. In that case, we have

and solving gives

, which makes more sense. Then the number of extra spades in the deck must be 2.
Zero. The negative one means arctan. Arctan times 0 is 0.
Answer:
width = 6
length = 4
Step-by-step explanation:
perimeter = 2L + 2W
L = W - 2
plug in perimeter:
20 = 2L + 2W
you know the length is 2 less than the width
plug in (w - 2) for L:
20 = 2(w - 2) + 2w
solve for w:
20 = 2w - 4 + 2w
20 = 4w - 4
4w = 24
w = 6
now plug in 6 for w in either equation, whichever is easier to solve:
L = 6 - 2
L = 4
width = 6
length = 4
Answer: i think its implied
Step-by-step explanation: