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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
a^7 - b^7 = (a - b)(a^6 + a^5.b + a^4.b^2 + a^3.b^3 + a^2.b^4 +a.b^5 + b^6)
(5*4)+(8*3) where *indicates to multiply, and the equations in parenthesis(5*4 and 8*3) are done before you add them, so 5*4=A, 8*3=B, then A+B. Not doing your homework for you. AND USE THE COMPUTER CALCULATOR!!
Multiplication is an abbreviation of addition: 4 x 5 is the same as: adding 4 together 5 times: 4 x 5 = 4 + 4 + 4 + 4 + 4 = 20; or adding 5 together 4 times: 4 x 5 = 5 + 5 + 5 + 5 = 20 When multiplying a negative number by a positive number, it can be thought of adding the negative number to itself the positive number of times. eg -4 x 5 = -4 + -4 + -4 + -4 + -4 = -20 eg -5 x 4 = -5 + -5 + -5 + -5 = -20 Adding a negative number is the same as subtracting the number (as a positive number): eg 7 + -3 = 7 - 3 = 4 eg -7 + -3 = -7 - 3 = -10
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
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.
This deals with ratios and proportions. ⊱ ────── ✯ ────── ⊰ 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 × 4 : 3 × 4 A : B = 8 : 12 Also, B : C = 4 × 3 : 5 × 3 B : C = 12 : 15 Now, we can combine A : B and B : C. A : B : C = 8 : 12 : 15.
I think you have not asked the question correctly.I guess you meant that the sides of the triangle are 3, 4 and 5. Similarly you have given no indication of which angle is opposite which side.A 3, 4, 5 triangle is a right angle triangle (5 is the hypotenuse).Thus depending where angle B is, its sine will be:If B is opposite the side of length 3, sin B = 4/5If B is opposite the side of length 4, sin B = 3/5If B is opposite the side of length 5, sin B = 5/5 = 1; alternatively ∠B = 90° and sin B = sin 90° = 1
5 [3-4-5 triangle]
a^7 - b^7 = (a - b)(a^6 + a^5.b + a^4.b^2 + a^3.b^3 + a^2.b^4 +a.b^5 + b^6)
You haven't provided any choices for the "which of the following" part of your question. Such questions are best avoided here. However, assuming a, b and c are all natural numbers, all of the following are true for a<b AND b+c=10: a=1, b=2, c=8 a=1, b=3, c=7 a=1, b=4, c=6 a=1, b=5, c=5 a=1, b=6, c=4 a=1, b=7, c=3 a=1, b=8, c=2 a=1, b=9, c=1 a=2, b=3, c=7 a=2, b=4, c=6 a=2, b=5, c=5 a=2, b=6, c=4 a=2, b=7, c=3 a=2, b=8, c=2 a=2, b=9, c=1 a=3, b=4, c=6 a=3, b=5, c=5 a=3, b=6, c=4 a=3, b=7, c=3 a=3, b=8, c=2 a=3, b=9, c=1 a=4, b=5, c=5 a=4, b=6, c=4 a=4, b=7, c=3 a=4, b=8, c=2 a=4, b=9, c=1 a=5, b=6, c=4 a=5, b=7, c=3 a=5, b=8, c=2 a=5, b=9, c=1 a=6, b=7, c=3 a=6, b=8, c=2 a=6, b=9, c=1 a=7, b=8, c=2 a=7, b=9, c=1 a=8, b=9, c=1
5 [3-4-5 triangle]
5 [3-4-5 triangle]
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
Given the legs a and b of a triangle are 3 and 4, the hypotenuse is: 5