1)√3 = 1.73 (near +2 left)
2)√10 = 3+something (near +3, towards RHS)
3)√26 = 5+something (near +5, towards RHS)
4)-√5 = -2-something (near -2 towards left)
5)-√14 = -4+something (near -4 towards right)
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.
The answer that i got was .163 because i divided the 1.63 dollars total by the total ounces (10) and got .163
The GCF of 16 and 40 is 8.
16/8 = 2
40/8 = 5
We can rewrite this using the distribute property like so:
<h3><u>8(2 + 5)</u></h3>
Using the distributive property:
16 + 40
16 + 40 = 56
We can also add inside the parentheses and multiply and we'll get the same answer.
8(7)
56
Answer:
cool
Step-by-step explanation:
whats ur question about it?