To solve this, you find out what 3m2= (3 meters squared). 3m2= 9, and 4m2= 16.
C plus is between 3 and 3.2. C = 75% 0% < Plus < 5% 75%+0% < C Plus < 75%+5% 75 < C Plus < 80% 75%*4 < C Plus < 80% * 4 (3/4)*4 < C Plus < (4/5) * 4 3 < C Plus < 16/5 3 < C Plus < 3.2
four a = 1 c = 3 a + c = 1 + 3 = 4
It is: 1-3(-4)+4(-2) = 5
Assuming the 10 = Cup A, 4 = Cup B and 3 = Cup C 1) Fill Cup C (A=0, B=0, C=3) 2) Pour Cup C into Cup A (A=3, B=0, C=0) 3) Fill Cup B (A=3, B=4, C=0) 4) Fill Cup C from Cup A (A=3, B=1, C=3) 5) Pour the remainder of Cup B into Cup A (A=4, B=0, C=3) 6) Empty Cup C (A=4, B=0, C=0) 7) Fill Cup B (A=4, B=4, C=0) 8) Fill Cup C from Cup A (A=4, B=1, C=3) 9) Pour the remainder of Cup B into Cup A (A=5, B=0, C=3) 10) Empty Cup C (A=5, B=0, C=0) 11) Fill Cup B (A=5, B=4, C=0) 12) Fill Cup C from Cup A (A=5, B=1, C=3) 13) Empty Cup C (A=5, B=1, C=0) 13) Pour the remainder of Cup B into Cup C (A=5, B=0, C=1) 14) Fill Cup B (A=5, B=4, C=1) so assuming you count the filling of cups as pours your answer is 14
-2
1 qt = 4 c 2 qt = 8 c 3 qt = 12 c 4 qt = 16 c
-2
You would do it it by oppisites. 3+2+4=9 which equals C.
The total number of outcomes you could get by flipping a coin 4 times is 2^4 or 16 ways as each coin toss yields two possible outcomes (Heads or Tails) and there are four trials. With that said, you need to find out how many ways there are to get 3 heads or 4 heads. You could use combinations to find this: n C r = n! / [ r! (n-r)! ] 4 C 3 = 4! / [ 3! (4-3)! ] 4 C 3 = 4! / 3! 4 C 3 = 4 4 C 4 = 4! / [ 4! (4-4)! ] 4 C 4 = 4! / 4! 4 C 4 = 1 The total numbers of ways that one could get at least 3 H heads is equal to five (4 ways to get three heads, and one way to get 4 heads). So the probability of getting at least 3 heads in four tosses is equal to 5/16 or 31.25%.
3.5 and a space :)
y = mx + c Here y = -4x + c Since it passes through -2, 3 3 = (-4)(-2) + c 3 = 8 + c 3 - 8 = c c = -5 Hence equation is y = -4x -5
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