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No.Neither are commutative: a - b does not equal b - a, and a/b does not equal b/a.Neither is associative: (a - b) - c does not equal a - (b - c), and (a/b)/c does not equal a/(b/c).Examples of these are:4 - 2 does not equal 2 - 4.1/3 does not equal 3/1.(6 - 5) - 1 does not equal 6 - (5 - 1).(10/2)/2 does not equal 10/(2/2).
1 is the answer. -(-4)- 3 = 4-3 = 1 .
3475636573626758335263257623587365
3 + 5 = 4*b so 8 = 4*b Divide both sides by 4: 2 = b
There are 20 unique 5-card hands. Solution: First break apart the player card combinations. You can either use player card 1, player card 2, or both player cards 1 and 2 with the hand. Therefore, you are taking 5 cards, 3 at a time, when playing with both player cards, and 5 cards, 4 at a time, when playing with 1 player card. Mathematically, this can be expressed using the Binomial coefficient 2 x ( 5 choose 4 ) + ( 5 choose 3 ) = Number of possible 5-card hands. 5 choose 4 = 5 : This represents the combinations between 1 player card and 5 table cards, 4 at a time. (http://www.wolframalpha.com/input/?i=5+choose+4) 5 choose 3 = 10 : This represents the combinations between 2 player cards, and 5 table cards, 3 at a time. (http://www.wolframalpha.com/input/?i=5+choose+3) Thus, 2 x ( 5 ) + ( 10 ) = 20 . Further Information: All possible hand combinations are listed below. A and B are the two player cards. 0 through 5 are the five table cards. ----- A 1 2 3 4 A 2 3 4 5 A 3 4 5 1 A 4 5 1 2 A 5 1 2 3 B 1 2 3 4 B 2 3 4 5 B 3 4 5 1 B 4 5 1 2 B 5 1 2 3 A B 2 3 4 A B 3 4 5 A B 1 2 3 A B 4 5 1 A B 5 1 2 A B 4 5 2 A B 4 1 2 A B 5 1 3 A B 5 2 3 A B 1 3 5
No because if a=4 and b=4 then a=b and not a can equal 6 and not b can equal 5 so ~a is not = to ~b
No.Neither are commutative: a - b does not equal b - a, and a/b does not equal b/a.Neither is associative: (a - b) - c does not equal a - (b - c), and (a/b)/c does not equal a/(b/c).Examples of these are:4 - 2 does not equal 2 - 4.1/3 does not equal 3/1.(6 - 5) - 1 does not equal 6 - (5 - 1).(10/2)/2 does not equal 10/(2/2).
given 1=52=253=3254=4325now the question is 5=?from basic mathematics we have if a=b then b=atherefore if 1=5 then 5=1.hence the answer is 1
1 is the answer. -(-4)- 3 = 4-3 = 1 .
This deals with ratios and proportions. โฑ โโโโโโ โฏ โโโโโโ โฐ A : B = 2 : 3 B : C = 4 : 5. Now, to find A : B : C, we need to make the value of B equal in A : B ratio and B : C ratio. Here, Value of B in A : B ratio is 3; and B : C ratio is 4. LCM of 3 and 4 is 12. Therefore, we multiply 4 to the first ratio and 3 to the second ratio. A : B = 2 ร 4 : 3 ร 4 A : B = 8 : 12 Also, B : C = 4 ร 3 : 5 ร 3 B : C = 12 : 15 Now, we can combine A : B and B : C. A : B : C = 8 : 12 : 15.
3475636573626758335263257623587365
its when you have a right triangle with lengths that equal 3, 4, and 5. A=3 B=4 C=5 C is your hypotenuse (sorry about spelling there)
Let's reverse the question - Is a over b less than a squared over b squared? Answer - Only when a is less than b example 1: a is less than b a = 2 a squared = 4 b = 3 b squared = 9 2 / 3 = .6666 4 / 9 = .444444 2 / 3 is greater than 4 / 9 example 2: a is equal to b a = 2 a squared = 4 b = 2 b squared = 4 2 / 4 = .5 2 / 4 = .5 2 / 4 is equal to 2 / 4 example 3: a is greater than b a = 3 a squared = 9 b = 2 b squared = 4 3 / 2 = 1.5 9 / 4 = 2.25 3 / 2 is less than 9 / 4 - wjs1632 -
3 + 5 = 4*b so 8 = 4*b Divide both sides by 4: 2 = b
Given that 8a + 7b + 3c = 3b + 6a + 6c, and a = 2/3c, and if 12b = 3, then b = 1/4 Substituting for a and b, 16/3c + 7/4 + 3c = 3/4 + 12/3c + 6c Making 12 the common denominator, 64/12c + 21/12 + 36/12c = 9/12 + 48/12c + 72/12c Combining like terms, 100/12c + 21/12 = 9/12 + 120/12c Grouping like terms, 12/12 = 20/12c Simplifying fractions, 1 = 5/3c Multiplying both terms by 5/3, c = 3/5 And a = 2/3c = 2/3 x 3/5 = 2/5 Checking the answer using a = 2/5, b =1/4, and c = 3/5, Left side = 8a + 7b + 3c = 16/5 + 7/4 + 9/5 = 25/5 + 7/4 = 5 + 1 3/4 = 6 3/4 Right side = 3b + 6a + 6c = 3/4 + 12/5 + 18/5 = 3/4 + 30/5 = 3/4 + 6 = 6 3/4
There are 20 unique 5-card hands. Solution: First break apart the player card combinations. You can either use player card 1, player card 2, or both player cards 1 and 2 with the hand. Therefore, you are taking 5 cards, 3 at a time, when playing with both player cards, and 5 cards, 4 at a time, when playing with 1 player card. Mathematically, this can be expressed using the Binomial coefficient 2 x ( 5 choose 4 ) + ( 5 choose 3 ) = Number of possible 5-card hands. 5 choose 4 = 5 : This represents the combinations between 1 player card and 5 table cards, 4 at a time. (http://www.wolframalpha.com/input/?i=5+choose+4) 5 choose 3 = 10 : This represents the combinations between 2 player cards, and 5 table cards, 3 at a time. (http://www.wolframalpha.com/input/?i=5+choose+3) Thus, 2 x ( 5 ) + ( 10 ) = 20 . Further Information: All possible hand combinations are listed below. A and B are the two player cards. 0 through 5 are the five table cards. ----- A 1 2 3 4 A 2 3 4 5 A 3 4 5 1 A 4 5 1 2 A 5 1 2 3 B 1 2 3 4 B 2 3 4 5 B 3 4 5 1 B 4 5 1 2 B 5 1 2 3 A B 2 3 4 A B 3 4 5 A B 1 2 3 A B 4 5 1 A B 5 1 2 A B 4 5 2 A B 4 1 2 A B 5 1 3 A B 5 2 3 A B 1 3 5
Inspection of the coordinates (by drawing a sketch) shows that the quadrilateral is a right trapezium with the parallel sides (a, b) horizontal. Let A = (-4, 3), B = (-4, 5), C = (3, 5), D = (5, 3) area trapezium = ½ × (a + b) × h = ½ × (BC + AD) × AB = ½ × ((3 - -4) + (5 - -4)) × (5 - 3) = ½ × (7 + 9) × 2 = 16 square units.