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Maria keeps her four stuffed bears lined up on a shelf over her bed how many arrangements of the bears are possible?
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
What is proof of Heron's Formula?
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)
Are there any online tools for converting math powers Eg something that can tell me 2 to the power of 4 equals 4 to the power of 2?
Your example is confusing -- while 2^4 = 4^2, in general it is false that x^y=y^x, so not sure what you mean by "converting powers." A relevant rule is that (a^b)^c = a^(bc) (power of a power is product of powers) If a=2, b=2, and c=2 then a^b = 2-squared = 4 and the formula gives 4^2 = 2^(2*2) which is true. Also (a^b)*(a^c) = a^(b+c) (multiply numbers, add powers) and (a^b)/(a^c) = a^(b-c) (divide numbers, subtract powers)
What does c mean in c x 2 equals b?
b divided by 2
What does a distributive property mean?
The distributive property states that a × (b + c) = a × b + a × c
What is 4 C B?
What is 4 C B
the ratio of A to B is 2:3 and the ratio of B to C is 4:5. what is the ratio of A to C?
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.
What 3 numbers have a mean of 10?
mean or average = (a + b + c) / 3 = 10 so a + b + c = 10 * 3, so a + b + c = 30, so take your pick, 2 + 10 + 18, 11 + 4 + 15 etc
What does a and b are factors of c mean?
It means a*b = c
What Does A 3.2 GPA Mean?
It equals to a low B 4 = A 3 = B 2 = C 1 = D 0 = F
If you have a 10 measuring cup 4 measuring cup and a 3 measuring cup how many pours does it take to get the 4 to 4 the 3 to 1 and the 10 to half?
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
What is A B C-BB?
Oh, and I mean A+B+C=BB
Maria keeps her four stuffed bears lined up on a shelf over her bed how many arrangements of the bears are possible?
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
What is proof of Heron's Formula?
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)
If A is less than B and B plus C equals 10 and none of them equal zero then which of the following must be true?
Well, isn't that just a happy little math problem! If A is less than B and B plus C equals 10, then it must be true that A plus C is less than 10. Just remember, in the world of numbers, everything adds up beautifully in the end.
Notes to play you got to feeling?
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 ;)
What prison tattoos are there in the UK and what do they symbolize?
4 dots, or A C A B which mean all cops are bastards
