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3 Quadratic Formula Finally, the quadratic formula if a, b and c are real numbers, then the quadratic polynomial equation ax2 bx c = 0 (31) has (either one or two) solutions x = b p b2 4ac 2a (32) 4 Points and Lines Given two points in the plane, P = (x 1;y 1);$$(a b c)^3 = (a^3 b^3 c^3) (3a^2b 3a^2c 3abc) (3ab^2 3b^2c 3abc) (3ac^2 3bc^2 3abc) 3abc$$ $$(a b c)^3 = (a^3 b^3 c^3) 3a(ab ac bc) 3b(ab bc ac) 3c(ac bc ab) 3abc$$There are n 4 – 1 C 4 – 1 = n 3 C 3 terms in the above expansion REMARK The greatest coefficient in the expansion of (x 1 x 2 x m ) n is (n!) / (q!) m – r {(q1)!} r , where q and r are the quotient and remainder, respectively when n is divided by m
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(a+b+c)^3 formula expansion
(a+b+c)^3 formula expansion-(say, a;b) to functions in the new variable z, whose domain is an interval of 2ˇlength This is because x= a)z= ˇ l aand x= b)z= ˇ l b= 2 b a (b a a) = 2ˇ ˇ l a Thus when the variable xin f(x) moves from ato b, the new variable zin the new function F(z) (which is the same function fin the new variable) moves from cto c2ˇ, where c= ˇ l a= 1 _3C_0 = (3!)/ ((30)!
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Play this game to review Algebra I Determine the values of a, b, and c for the quadratic equation 4x 2 – 8x = 3 Preview this quiz on Quizizz Determine the values of a, b, and c for the quadratic equation 4x2 – 8x = 3= (5)(4)(3)(2)(1) = 1` (c) `(4!)/(2!)=((4)(3)(2)(1))/((2)(1))=12` Note `(4!)/(2!)` cannot be cancelled down to `2!` c Factorial Interactive Instructions You can use the following interactive to find the factorial of any positive integer up to 30!A^3 b^3 c^3 3abc = (a b c) (a^2 b^2 c^2 ab ac bc) By assumption a^3b^3c^3=3abc so the left hand side is 0 Therefore (abc) (a^2b^2c^2abacbc) = 0 So either abc=0 or
1 decade ago Favorite Answer = (a b) (a b) (a b) = (a b) (a² ab ab b²) = (a b) (a² 2ab b²) = a³ 2a²b ab² a²b 2ab² b³ = a³ 3a²b 3ab² b³ Answer a³ 3a²b 3ab² b³We can do so in two waysWhat Is The Expansion Of A B C 3 Quora For more information and source, see on this link https//wwwquoracom/Whatistheexpansionofabc3
A³ a²b a²c 2a²b 2ab² 2abc 2a²c 2abc 2ac² ab² b³ b²c 2abc 2b²c 2bc² ac² bc² c³ This is just multiplying out and bookkeeping It's a^3 b^3 c^3 plus 3 of each= 2*1 = 2, 1!1 decade ago Favorite Answer = (a b) (a b) (a b) = (a b) (a² ab ab b²) = (a b) (a² 2ab b²) = a³ 2a²b ab² a²b 2ab² b³ = a³ 3a²b 3ab² b³ Answer a³ 3a²b 3ab² b³
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= 1 and 0!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) 5Note again The lower index, in this case 4, is the exponent of b This same number is also the coefficient of a 4 b 8, since 12 C 8 = 12 C 4 Example 2 Expand (a − b) 5 SolutionWe found the binomial coefficients to be 1 5 10 10 5 1 The difference with (a − b) is that the signs of the terms will alternate(a − b) 5 = a 5 − 5a 4 b 10a 3 b 2 − 10a 2 b 3 5ab 4 − b 5
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\(=> (abc)^3 = a^3 b^3 c^3 \\ 6abc 3a^2b 3ab^2 \\ 3ac^2 3bc^2 3b^2c 3a^2c \) \(=> (abc)^3 = \\a^3 b^3 c^3 6abc 3ab (ab) 3ac (ac) 3bc (bc) \) (abc)^3 Verifications Need to verify \( (a b c)^ 3 \) formula is right or wrong put the value of a = 1, b=2 and c=3 put the value of a and b in the LHSFree math lessons and math homework help from basic math to algebra, geometry and beyond Students, teachers, parents, and everyone can find solutions to their math problems instantlyLet's explore the coefficients further Suppose that we want to find an expansion of (a b) 6 The patterns we just noted indicate that there are 7 terms in the expansion a 6 c 1 a 5 b c 2 a 4 b 2 c 3 a 3 b 3 c 4 a 2 b 4 c 5 ab 5 b 6 How can we determine the value of each coefficient, c i?
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In binomial theorem expansion, the binomial expression is most important in an algebraic equation which holds two different terms Such as a b, a3 b3, etc Let's consider;Explanation Binomial formula for (a b)3 ⇒3 C0a3b0 3 C1a2b1 3 C2a1b2 3 C3a0b3 Here, a = x and b = 1 ⇒3 C0x3 3 C1x2 × 11 3 C2x1 ×12 3 C3 × 13 As →3 C0 =3 C3 = 1 and →3 C1 =3 C2 = 3 ⇒ x3 3x2 3x 1 Answer linkWhat I want to do with this video is cover something called the triple product expansion or Lagrange's formula, sometimes And it's really just a simplification of the cross product of three vectors, so if I take the cross product of a, and then b cross c
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$$ \fbox{ Triangle 3 } \\ \red a^2 = b^2 c^2 2bc \cdot cos (A) \\ \red a^2 = 185^2 16^2 2\cdot 185 \cdot 16 \cdot cos (\red A) $$ Since we don't know the included angle, $$ \angle A $$ , our formula does not helpwe end up with 1 equation and 2 unknownsLinear thermal expansion is ΔL = αLΔT, where ΔL is the change in length L, ΔT is the change in temperature, and α is the coefficient of linear expansion, which varies slightly with temperature The change in area due to thermal expansion is Δ A = 2α A Δ T , where Δ A is the change in areaN ∈ N Then the result will be ∑ i = 0 n n C r x n − r y r n C r x n − r y r n C n − 1 x y n − 1 n C n y n
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Solve your math problems using our free math solver with stepbystep solutions Our math solver supports basic math, prealgebra, algebra, trigonometry, calculus and moreQ = (x 2;y 2) you can obtain the following information 1The distance betweenA 3 b 3 c 3 3abc= (ab) 3 3a 2 b3ab 2 c 3 3abc = (abc) 3 3 (ab) 2 c3 (ab)c 2 3ab (abc) = (abc) 3 3 (ab)c (abc)3ab (abc)= (abc) (a 2 b 2 c 2 abbcac) Cite 4
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(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 4In Algebra In Algebra putting two things next to each other usually means to multiply So 3 (ab) means to multiply 3 by (ab) Here is an example of expanding, using variables a, b and c instead of numbers And here is another example involving some numbers Notice the "·" between the 3 and 6 to mean multiply, so 3·6 = 18When the temperature is 22 °C, there is a gap of 10 x 103 m separating their ends No expansion is possible at the other end of either rod At what temperature will the two bars touch?
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= 3*2*1 = 6, 2!The change in temperature is the same for both, the original length is known for both, and the coefficients of linear expansion can be found from Table 122(0!)) = (3!)/ ((3)!1) = 1
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4 Binomial Expansions 41 Pascal's riTangle The expansion of (ax)2 is (ax)2 = a2 2axx2 Hence, (ax)3 = (ax)(ax)2 = (ax)(a2 2axx2) = a3 (12)a 2x(21)ax x 3= a3 3a2x3ax2 x urther,F (ax)4 = (ax)(ax)4 = (ax)(a3 3a2x3ax2 x3) = a4 (13)a3x(33)a2x2 (31)ax3 x4 = a4 4a3x6a2x2 4ax3 x4 In general we see that the coe cients of (a x)n come from the nth row of Pascal'sN ∈ N Then the result will be ∑ i = 0 n n C r x n − r y r n C r x n − r y r n C n − 1 x y n − 1 n C n y nIt is sometimes useful to write the cube of a binomial as follows 31 For instance, A=a, and B=bc 32 Expanding the middle part first 33 Finally expanding (bc) 3 34 Combinational Approach for Cubes
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Geometrically the trivector a ∧ b ∧ c corresponds to the parallelepiped spanned by a, b, and c, with bivectors a ∧ b, b ∧ c and a ∧ c matching the parallelogram faces of the parallelepiped As a trilinear functional The triple product is identical to the volume form of the Euclidean 3space applied to the vectors via interior productThere are other ways of solving a quadratic equation instead of using the quadratic formula, such as factoring (direct factoring, grouping, AC method), completing the square, graphing and others Given a general quadratic equation of the form ax²bxc=0 with x representing an unknown, a, b and c representing constants with a ≠ 0, theIn binomial theorem expansion, the binomial expression is most important in an algebraic equation which holds two different terms Such as a b, a3 b3, etc Let's consider;
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3(a b)3 = a3 b3 3ab(a b) 4 (a b) 3 = a3 b3 3ab(a b) 5(a b c)2 = a2 b2 c2 2ab2bc 2ca 6(a b c)3 = a3 b3 c33a2b3a2c 3b2c 3b2a 3c2a 3c2a6abc 7a2 b2 = (a b)(a – b ) 8a3 – b3 = (a – b) (a 2 ab b2 ) 9a3 b3 = (a b) (a 2 ab b2 ) 10(a b)2 (a b) 2 = 4ab 11(a b)2 (a b) 2 = 2(a 2 b2 )A1/3 a1/3 a1/3 = a (24) (a1/3)3 = a (25) (a2)1/3 = (a1/3)2 = a2∕3 (26) (a1/3)1/4 = a1/3 1/4 = (a1/4)1/3 (27) (a b)1/3 = a1/3 b1/3 (28) (a / b)1/3 = a1/3 / b1/3 (29) (1 / a)1/3 = 1 / a1/3 = a1/3 (30) Sponsored Links Mathematics Mathematical rules and laws numbers, areas, volumes, exponents, trigonometric functions and moreX, y ∈ R;
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The Formula is given below (a b c)³ = a³ b³ c³ 3 (a b) (b c) (a c) Explanation Let us just start with (abc)² = a² b² c²2ab2bc2ca =a² b² c²2 (abbcca) now (abc)² (abc)= (a b c)³= a² b² c²2 (abbcca) (abc) =a ² abc b² abcc² abc 2 (abbcca) abcSo, the value of a 2 b 2 c 2 is 175 Example 3 If a b c = 36 and a 2 b 2 c 2 = 676, then find the value of (ab bc ca) Solution To get the value of (ab bc ac), we can use the formula or expansion of (a b c) 2 Write the formula / expansion for (a b c) 2 (a b c) 2 = a 2 b 2 c 2 2ab 2bc 2acA^3 b^3 c^3 3abc = (a b c) (a^2 b^2 c^2 ab ac bc) By assumption a^3b^3c^3=3abc so the left hand side is 0 Therefore (abc) (a^2b^2c^2abacbc) = 0 So either abc=0 or
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X, y ∈ R;A 3 b 3 c 3 − 3abc = (a b c) (a 2 b 2 c 2 − ab − bc − ac) If (a b c) = 0, a 3 b 3 c 3 = 3abcA³ 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 questions
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A^3 b^3 c^3 plus 3 of each term having one variable and another one squared like ab^2, b^2c, all 6 combinations of those, then plus 6abc and that's itIn mathematics, a trinomial expansion is the expansion of a power of a sum of three terms into monomials The expansion is given by The expansion is given by ( a b c ) n = ∑ i j k = n i , j , k ( n i , j , k ) a i b j c k , {\displaystyle (abc)^{n}=\sum _{\stackrel {i,j,k}{ijk=n}}{n \choose i,j,k}\,a^{i}\,b^{\;\!j}\;\!c^{k},}3(a b)3 = a3 b3 3ab(a b) 4 (a b) 3 = a3 b3 3ab(a b) 5(a b c)2 = a2 b2 c2 2ab2bc 2ca 6(a b c)3 = a3 b3 c33a2b3a2c 3b2c 3b2a 3c2a 3c2a6abc 7a2 b2 = (a b)(a – b ) 8a3 – b3 = (a – b) (a 2 ab b2 ) 9a3 b3 = (a b) (a 2 ab b2 ) 10(a b)2 (a b) 2 = 4ab 11(a b)2 (a b) 2 = 2(a 2 b2 )
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So the answer is 3 3 3 × (3 2 × x) 3 × (x 2 × 3) x 3 (we are replacing a by 3 and b by x in the expansion of (a b) 3 above) Generally It is, of course, often impractical to write out Pascal"s triangle every time, when all that we need to know are the entries on the nth line(a 3 3a 2 b 3ab 2 b 3)(ab) = a 4 4a 3 b 6a 2 b 2 4ab 3 b 4 The calculations get longer and longer as we go, but there is some kind of pattern developing That pattern is summed up by the Binomial TheoremThe scalar triple product is unchanged under a circular shift of its three operands (a, b, c) ⋅ (×) = ⋅ (×) = ⋅ (×)
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= (3)(2)(1) = 6` (b) `5!(b)If exactly one member must be a woman, there are C(3;1)C(16;3) = 3560 = 1680 ways to form the committee (c)All of the committees formed in (b) qualify except those in which Smith and Jones are both members There are C(3;1)C(14;1) = 42 committees that both Smith and Jones can serve on, so C(3;1)C(16;3) C(3;1)C(14;1) = 1680 42 = 1638 committees canCopper (m/m o C) (17 μm/m o C) more coefficients more metals Example Expansion of an Aluminum Beam An aluminum construction is designed for temperatures ranging 30 o C to 50 o C If a beam's length is 6 m when assembled at o C the shortest final length of the beam at minimum temperature 30 o C can be calculated as
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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;We will construct this expansion in steps, and in so doing, derive the multinomial theorem Start with three nested summations The problem is that this triple summation produces 5·5·5=125 terms of various degrees ranging from a 0 b 0 c 0 to a 4 b 4 c 4According to this theorem, it is possible to expand the polynomial \((x y)^n\) into a series of the sum involving terms of the form a \(x^b y^c\) Here the exponents b and c are nonnegative integers with condition that b c = n Also, the coefficient of each term is a specific positive integer depending on n and b For example (for n = 4), we have \((xy)^4=x^44x^3y6x^2y^24xy^3y^4\) It is obvious that such expressions and their expansions would be very painful to multiply out by hand
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2 29 if a ib=0 wherei= p −1, then a= b=0 30 if a ib= x iy,wherei= p −1, then a= xand b= y 31 The roots of the quadratic equationax2bxc=0;a6= 0 are −b p b2 −4ac 2a The solution set of the equation is (−b p 2a −b− p 2a where = discriminant = b2 −4ac 32Solution This proceeds as Given polynomial (8a 3 27b 3 125c 3 – 90abc) can be written as (2a) 3 (3b) 3 (5c) 3 – 3 (2a) (3b) (5c) And this represents identity 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 = 5cUse the Binomial expansion (note the exponents sum to the power in each term) (xy)^3 = _3C_0x^3y^0 _3C_1x^2y^1 _3C_2x^1y^2 _3C_3x^0y^3 Remember 3!
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