What is 4 C B
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.
5 its 4
"6" It all depends on the relationship between A B and C. EG if a+b=c the c=6, if axb=c then c=8. you can throw in allsorts of relationships C/A = B So more information is required in the question
I assume you meant "sum". The answer is 60. 60=4 (integers) * 15 (average value of the integers). average of 4 integers = 15 --> (a + b + c + d)/4 = 15 --> (a + b + c + d) = 15 * 4 --> (a + b + c + d) = 60
1.b 2.a 3.a 4.b 5.c 6.c 7.c 8.b
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
This is a proof that uses the cosine rule and Pythagoras' theorem. As on any triangle with c being the opposite side of θ and a and b are the other sides: c^2=a^2+b^2-2abcosθ We can rearrange this for θ: θ=arccos[(a^2+b^2-c^2)/(2ab)] On a right-angle triangle cosθ=a/h. We can therefore construct a right-angle triangle with θ being one of the angles, the adjacent side being a^2+b^2-c^2 and the hypotenuse being 2ab. As the formula for the area of a triangle is also absinθ/2, when a and b being two sides and θ the angle between them, the opposite side of θ on the right-angle triangle we have constructed is 4A, with A being the area of the original triangle, as it is 2absinθ. Therefore, according to Pythagoras' theorem: (2ab)^2=(a^2+b^2-c^2)^2+(4A)^2 4a^2*b^2=(a^2+b^2-c^2)^2+16A^2 16A^2=4a^2*b^2-(a^2+b^2-c^2)^2 This is where it will start to get messy: 16A^2=4a^2*b^2-(a^2+b^2-c^2)(a^2+b^2-c^2) =4a^2*b^2-(a^4+a^2*b^2-a^2*c^2+a^2*b^2+b^4-b^2*c^2- a^2*c^2-b^2*c^2+c^4) =4a^2*b^2-(a^4+2a^2*b^2-2a^2*c^2+b^4-2b^2*c^2+c^4) =-a^4+2a^2*b^2+2a^2*c^2-b^4+2b^2*c^2-c^4 (Eq.1) We will now see: (a+b+c)(-a+b+c)(a-b+c)(a+b-c) =(-a^2+ab+ac-ab+b^2+bc-ac+bc+c^2)(a^2+ab-ac-ab-b^2+bc+ac+bc-c^2) =(-a^2+b^2+2bc+c^2)(a^2-b^2+2bc-c^2) =-a^4+a^2*b^2-2a^2*bc+a^2*c^2+a^2*b^2-b^4+2b^3*c-b^2*c^2+2a^2*bc-2b^3*c+(2bc)^2-2bc^3+a^2*c^2-b^2*c^2+2bc^3-c^4 =-a^4+2a^2*b^2+2a^2*c^2-b^4+(2bc)^2-c^4-2b^2*c^2 =-a^4+2a^2*b^2+2a^2*c^2-b^4+2b^2*c^2-c^4 (Eq.2) And now that we know that Eq.1=Eq.2, we can make Eq.1=(a+b+c)(-a+b+c)(a-b+c)(a+b-c) Therefore: 16A^2=(a+b+c)(-a+b+c)(a-b+c)(a+b-c) A^2=(a+b+c)(-a+b+c)(a-b+c)(a+b-c)/16 =[(a+b+c)/2][(-a+b+c)/2][(a-b+c)/2][(a+b-c)/2] And so if we let s=(a+b+c)/2 A^2=s(s-a)(s-b)(s-c)
The answer is 4! (4 factorial), the same as 4x3x2x1, which equals 24 combinations. The answer is 24 and this is how: A b c d A b d c A c d b A c b d A d c b A d b c B c d a B c a d B d a c B d c a B a c d B a d c C d a b C d b a C a b d C a d b C b d a C b a d D a b c D a c b D b c a D b a c D c a b D c b a
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
here's the notes for i gotta feeling on the recorder by the black eyed peas: d' d' d' d' c' c' bb c'c'c'c' c'c'c'c' b b b b b b b b a ( 4 beats) g ( 4 beats) a ( 4 beats) g ( 4 beats) c' a ( 4 beats) g ( 4 beats) a ( 4 beats) g ( 4 beats) E A G C' B..........E E C' B A G E C' B ......... E E C' B A G E D' B E E D' B A G E D' D' B A G ... C' B E E C' B A G E C' B ......... E E C' B A G E D' B E E D' B A G E D' D' B A G ... C' B E E C' B A G E C' B ......... E E C' B A G E D' B E E D' B A G E D' D' B A G ... C' B E E C' B A G E C' B ......... E E C' B A G E D' B E E D' B A G E D' D' B A G ... C' B A...G. Thats the begining thnks ;)
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.
Here's the notes for i gotta feeling on the recorder ( can also be played on the keyboard with the same notes) by the black eyed peas:d' d' d' d' c' c' bb c'c'c'c' c'c'c'c' b b b b b b b b c'c'c'c' c'c'c'c'a ( 4 beats) g ( 4 beats)a ( 4 beats) g ( 4 beats) c'a ( 4 beats) g ( 4 beats)a ( 4 beats) g ( 4 beats)E A G C' B..........E E C' B A G E C' B ......... E E C' B A G E D' B E E D' B A G E D' D' B A G ... C' BE E C' B A G E C' B ......... E E C' B A G E D' B E E D' B A G E D' D' B A G ... C' BE E C' B A G E C' B ......... E E C' B A G E D' B E E D' B A G E D' D' B A G ... C' BE E C' B A G E C' B ......... E E C' B A G E D' B E E D' B A G E D' D' B A G ... C' BA...G. Thats the begining thnks ;)
a+(b+c)=b+(a+c) 2+(7+4)=7+(2+4)
PLEASE NOTE ~ |= MEASURE SEPARATION ALL OF THE Ds ARE HIGH D AND OPEN D WILL BE WRITTEN IN ITALICS ( D )4/4 B B B B B B| B D G A B| C# C# C# C# C# B B|B A A B D|B B B B B B| B D G A B| C# C# C# C# C# B B B| D D C# A G| D B A G D | D B A G E | E C# B A F | D D C# A B| D B A G D | D B A G E | E C# B A D D D D | E-(high) D C# A G D| B B B B B B| B D G A B| C# C# C# C# C# B B|B A A B D| B B B B B B| B D G A B| C# C# C# C# C# B B B| D D C# A G|
1.A 2.d 3.b 4.a 5.c 6.a 7.c 8.c 9.b 10.c 11.b 12.b 13.a 14.a 15.a
It helps to consider subtraction as the addition of the additive inverse, in this case, a - b = a + (-b), where (-b) is the additive inverse of b. In this case: (a - b) - c = (a + (-b)) + (-c) = a + (-b) + (-c) a - (b - c) = a + -(b + (-c)) = a + -(b + (-c)) = a + -b + c As you can see, the "c" part is inverted. Here is an example with numbers: (10 - 5) - 1 = 4 10 - (5 - 1) = 10 - 4 = 6 In the last subtraction, the result is that the 1 is added instead of subtracted.
3.2