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Using determinants, we can write the result as a × b = a 2 a 3 b 2 b 3 i − a 1 a 3 b 1 b 3 j a 1 a 2 b 1 b 2 k Looking at the formula for the 3 × 3 determinant, we see that the formula for a cross product looks a lot like the formula for the 3 × 3 determinantExample 3 A = {a, b, c, d, e}, B = {x, y, z} and C = {a, e, x} find the value of n(AUBUC) Solution n(AUBUC)=n(A)n(B) n(C)n(AnB)n(BnC)n(CnA)n(AnBnC) n (A) = 5 n (B) = 3 n (C) = 3 A n B = {a, b, c, d, e} n {x, y, z} = { } n (A n B) = 0 B n C = {x, y, z} n {a, e, x} = {x} n (B n C) = 1The Triangle Formula are given below as, Perimeter of a triangle = a b c \Area\;
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(a+b+c)2 ka formula
(a+b+c)2 ka formula-A 3 b 3 = (a b) (a 2 b 2 − ab) (a b c) 3 = a 3 b 3 c 3 3(a b)(b c)(c a) a 3 b 3 c 3 − 3abc = (a b c) (a 2 b 2 c 2 − ab − bc − ac)Solving for B means to isolate it to one side of the equation, doesn't matter which side A= ABC/3 Start by subtracting A from both sides
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MATHEMATICAL FORMULAE Algebra MATHEMATICAL FORMULAE Algebra 1 (ab)2=a22abb2;a2b2=(ab)2−2ab 2 (a−b)2=a2−2abb;a2b2=(a−b)22ab 3 (abc)2=a2b2c22(abbcca) 4 (ab)3=a3b33ab(ab);a3b3=(ab)−3ab(ab) 5(a b c) 2 = a b (c) 2 = a 2 b 2 (c) 2 2ab 2 (b) (c) 2 (c) (a) = a 2 b 2 c 2 2ab – 2bc 2ca Therefore, (a b c) 2 = a 2 b 2 c 2 2ab – 2bc 2ca (a b c) 2 = a ( b) c 2 = a 2 (b 2) c 2 2 (a) (b) 2 (b) (c) 2 (c) (a) = a 2 b 2 c 2 – 2ab – 2bc 2ca Therefore, (a b c) 2 = a 2 b 2 c 2 – 2ab – 2bc 2ca(a b c)³ = a³ b³ c³ 3 (a b) (b c) (a c) It can be written as (a b c)³ a³ b³ c³ = 3 (a b) (b c) (a c) (1) Consider the LHS of equation (1), (a b c)³ a³ b³ c³ = a³ b³ c³ 3 ab (a b) 3 bc (b c) 3 ac (a c) 6 abc a³ b³ c³
The formula for Solution To find the formula for Multiply twice and simplify the terms Squaring the whole equation, we can see that the value of (abc)^{2} is Taking 2 common in we get Therefore, the formula for is This formula is used to express the algebraic identity if they are 3 terms given according to the operations involvedThe Triangle Formula are given below as, Perimeter of a triangle = a b c \Area\;(a b) 3 = a 3 b 3 3ab(a b) (a – b) 3 = a 3 – 3a 2 b 3ab 2 – b 3 = a 3 – b 3 – 3ab(a – b) a 3 – b 3 = (a – b)(a 2 ab b 2) a 3 b 3 = (a b)(a 2 – ab b 2) (a b) 4 = a 4 4a 3 b 6a 2 b 2 4ab 3 b 4
A 3 b 3 c 3 3abc = (a b c)(a 2 b 2 c 2 ab bc ca) Where a = 2a, b = 3b and c = 5c Now apply values of a, b and c on the LHS of identity ie a 3 b 3 c 3 3abc = (a b c)(a 2 b 2 c 2 ab bc ca) and we get = (2a 3b 5c) { (2a) 2 (3b) 2 (5c) 2 – (2a)(3b) – (3b)(5c) – (5c)(2a) } Expand the exponential forms and we getLooking at the formula for the $3 \times 3$ determinant, we see that the formula for a cross product looks a lot like the formula for the $3 \times 3$ determinant If we allow a matrix to have the vector $\vc{i}$, $\vc{j}$, and $\vc{k}$ as entries (OK, maybe this doesn't make sense, but this is just as a tool to remember the cross product), theOnly the use of the quadratic formula, as well as the basics of completing the square will be discussed here (since the derivation of the formula involves completing the square) Below is the quadratic formula, as well as its derivation Derivation of the Quadratic Formula From this point, it is possible to complete the square using the relationship that x 2 bx c = (x h) 2 k Continuing the derivation using this relationship
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There are mainy two forms to write the formula of a 3 b 3 c 3 First one is a 3 b 3 c 3 = ( a b c) a 2 b 2 c 2 − a b − b c − a c 3 a b c Second one is a 3 b 3 c 3 = ( a b c) 1 2 ( a − b) 2 ( b − c) 2 ( c − a) 2 3 a b c(a b c) 2 = a b (c) 2 = a 2 b 2 (c) 2 2ab 2 (b) (c) 2 (c) (a) = a 2 b 2 c 2 2ab – 2bc 2ca Therefore, (a b c) 2 = a 2 b 2 c 2 2ab – 2bc 2ca (a b c) 2 = a ( b) c 2 = a 2 (b 2) c 2 2 (a) (b) 2 (b) (c) 2 (c) (a) = a 2 b 2 c 2 – 2ab – 2bc 2ca Therefore, (a b c) 2 = a 2 b 2 c 2 – 2ab – 2bc 2ca(abc) 3 a 3 b 3 c 3 We can choose three "a"'s for the cube in one way C(3,3)=1, or we can choose an a from the first factor and one from the second and one from the third, being the only way to make a3 The coefficient of the cubes is therefore 1 (It's the same for a, b and c, of course) 3a 2 b3a 2 c Next, we consider the a 2 terms We can choose two a's from 3 factors in C(3,2) ways=3
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A= ABC/3 Answer by nerdybill (7384) ( Show Source ) You can put this solution on YOUR website!As stated in the title, I'm supposed to show that (a b c) 3 = a 3 b 3 c 3 (a b c) (a b a c b c)Only the use of the quadratic formula, as well as the basics of completing the square will be discussed here (since the derivation of the formula involves completing the square) Below is the quadratic formula, as well as its derivation Derivation of the Quadratic Formula From this point, it is possible to complete the square using the relationship that x 2 bx c = (x h) 2 k Continuing the derivation using this relationship
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Video Lesson Khan Academy Video Quadratic Formula 1;Binomial Theorem (ab)1 = a b ( a b) 1 = a b (ab)2 = a2 2abb2 ( a b) 2 = a 2 2 a b b 2 (ab)3 = a3 3a2b 3ab2 b3 ( a b) 3 = a 3 3 a 2 b 3 a b 2 b 3 (ab)4 = a4 4a3b 6a2b2 4ab3 b4 ( a b) 4 = a 4 4 a 3 b 6 a 2 b 2 4 a b 3 b 4Example 3 A = {a, b, c, d, e}, B = {x, y, z} and C = {a, e, x} find the value of n(AUBUC) Solution n(AUBUC)=n(A)n(B) n(C)n(AnB)n(BnC)n(CnA)n(AnBnC) n (A) = 5 n (B) = 3 n (C) = 3 A n B = {a, b, c, d, e} n {x, y, z} = { } n (A n B) = 0 B n C = {x, y, z} n {a, e, x} = {x} n (B n C) = 1
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(a b c) 2 = a 2 b 2 c 2 2ab 2bc 2ca (a – b – c) 2 = a 2 b 2 c 2 – 2ab 2bc – 2ca (a b) 3 = a 3 3a 2 b 3ab 2 b 3;Triangle= \frac{1}{2}bh\ Where, b is the base of the triangle h is the height of the triangle If only 2 sides and an internal angle is given then the remaining sides and angles can be calculated using the below formulaSolve an equation of the form a x 2 b x c = 0 by using the quadratic formula x = − b ± √ b 2 − 4 a c 2 a StepByStep Guide Learn all about the quadratic formula with this stepbystep guide Quadratic Formula, The MathPapa Guide;
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A 3 b 3 c 3 3abc = (a b c)(a 2 b 2 c 2 ab bc ca) Where a = 2a, b = 3b and c = 5c Now apply values of a, b and c on the LHS of identity ie a 3 b 3 c 3 3abc = (a b c)(a 2 b 2 c 2 ab bc ca) and we get = (2a 3b 5c) { (2a) 2 (3b) 2 (5c) 2 – (2a)(3b) – (3b)(5c) – (5c)(2a) } Expand the exponential forms and we get(abc)^3 Formula A Plus B Plus C Whole Square (abc)^3 Proof = a^3 b^3 c^3 6abc 3ab (ab) 3ac (ac) 3bc (bc)Determine the values of a, b, and c for the quadratic equation 4x 2 – 8x = 3 answer choices a = 4, b = 8, c = 3 a = 4, b =8, c =3 a = 4, b = 8, c = 3 a = 4, b = 8, c = 3 s Question 2
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The plural of formula can be either formulas or, under the influence of scientific Latin, formulae In mathematics, a formula generally refers to an identity which equates one mathematical expression to another, with the most important ones being mathematical theorems Syntactically, a formula is an entity which is constructed using the symbols and formation rules of a given logical language For example, determining the volume of a sphere requires a significant amount of integral calculus or itA 3 b 3 c 3 – 3abc = (a b c)(a 2 b 2 c 2 – ab – bc – ca) If a b c = 0, then the above identity reduces to a 3 b 3 c 3 = 3abc Few Other Mathematical FormulaSay, you're typing your letter in C1 and want it to appear in D1 In D1, use the formula =VLOOKUP (C1,AB,2,0) Breaking this down, it means C1 Take the value in C1 AB Look for it in the array composed of columns A trough B 2 Find the match in the 1st column and return the value that's in the 2nd column
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Heron's Formula is used to calculate the area of a triangle with the three sides of the triangle You have to first find the semiperimeter of the triangle with three sides and then area can be calculated based on the semiperimeter of the triangle The formula was derived by Hero of Alexendria, a Greek Engineer and Mathematician(abc) 3 a 3 b 3 c 3 We can choose three "a"'s for the cube in one way C(3,3)=1, or we can choose an a from the first factor and one from the second and one from the third, being the only way to make a3 The coefficient of the cubes is therefore 1 (It's the same for a, b and c, of course) 3a 2 b3a 2 c Next, we consider the a 2 terms We can choose two a's from 3 factors in C(3,2) ways=3Determine the values of a, b, and c for the quadratic equation 4x 2 – 8x = 3 answer choices a = 4, b = 8, c = 3 a = 4, b =8, c =3 a = 4, b = 8, c = 3 a = 4, b = 8, c = 3 s Question 2
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A³ b³ = (a b)(a² – ab b²) you know that (a b)³ = a³ 3ab(a b) b³ then a³ b³ = (a b)³ – 3ab(a b) = (a b)(a b)² – 3ab = (a b)(a² 2ab b² – 3ab) = (a b)(a² – ab b² ) Please log inor registerto add a comment Related questionsAnd we must go thru the formulae sequentially a (BC)2 ≡ ()2(B5C2−)2 the charge on this species is 10 so this is not a contender b A3(BC4)2 ≡ ()3(B5C2− 4)2 the charge on this species is 0 so this is a contender${\left( {a b c} \right)^2} = {a^2} {b^2} {c^2} 2ab 2ac 2bc$ ${\left( {a b} \right)^3} = {a^3} 3{a^2}b 3a{b^2} {b^3};{\left( {a b} \right)^3} = {a^3} {b^3} 3ab\left( {a b} \right)$ ${\left( {a b} \right)^3} = {a^3} 3{a^2}b 3a{b^2} {b^3}$ ${a^3} {b^3} = \left( {a b} \right)\left( {{a^2} ab {b^2}} \right)$
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4 years ago A B C 3 Formula Source (s) https//shrinkurlim/badse DanielM Lv 4 1 decade ago This is just multiplying out and bookkeeping It's a^3 b^3 c^3 plus 3 of each term having oneTriangle= \frac{1}{2}bh\ Where, b is the base of the triangle h is the height of the triangle If only 2 sides and an internal angle is given then the remaining sides and angles can be calculated using the below formula${\left( {a b c} \right)^2} = {a^2} {b^2} {c^2} 2ab 2ac 2bc$ ${\left( {a b} \right)^3} = {a^3} 3{a^2}b 3a{b^2} {b^3};{\left( {a b} \right)^3} = {a^3} {b^3} 3ab\left( {a b} \right)$ ${\left( {a b} \right)^3} = {a^3} 3{a^2}b 3a{b^2} {b^3}$ ${a^3} {b^3} = \left( {a b} \right)\left( {{a^2} ab {b^2}} \right)$
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