GCF(a2b3, ab3) = ab3
54 = 2 x 3 x 3 x 3 = 2 x 33. Thus if ab3 = 54 and a, b are prime, then a=2, b=3.
The GCF for 16 24 18 is 2.
The GCF is: 1
The GCF is 2x.
The GCF is xy
a2b3
Trygonal pyramid
1000 pounds
54 = 2 x 3 x 3 x 3 = 2 x 33. Thus if ab3 = 54 and a, b are prime, then a=2, b=3.
To find the cross product of two vectors:If a = [a1, a2, a3] and b = [b1, b2, b3], thena x b = [a2b3 - a3b2, a3b1 - a1b3, a1b2 - a2b1] or(a2b3 - a3b2)i + (a3b1 - a1b3)j + (a1b2 - a2b1)k
a5+b5 = (a+b) (a4-a3b+a2b2-ab3+b4)
because the dipoles changes from different AB3 molecule and the change of the bonding electrons pairs and the lone electrons pairs. eg. BF3 has (3BP) the shape is trigonal planar PCl3 has (3BP and 1LP) the shape is trigonal pyramidal BrF3 has (3BP and 2Lp) the shape is T-shaped
Assuming that you want (ab^3)^4, which is impossible to ask given the crap browser used by Answers, the solution is A^4b^12.
The chemical formula will be A2B3.
Prove that (axb)n[(bxc)x(cxa)] = [a]n(bxc)]^2 where a,b,and c are all vectors. First, multiply out the cross products. Since the cross product of two vectors is itself a vector, we'll give the cross products some names to make this a little easier to understand: (bxc)=(b2c3-b3c2)i-(b1c3-b3c1)j+(b1c2-b2c1)k = vector d (cxa)=(c2a3-c3a2)i-(c1a3-c3a1)j+(c1a2-c2a1)k = vector v (axb)=(a2b3-a3b2)i-(a1b3-a3b1)j+(a1b2-a2b1)k = vector u => (axb)n[(bxc)x(cxa)] = un[dxv] (dxv)=(d2v3-d3v2)i-(d1v3-d3v1)j+(d1v2-d2v1)k = vector w => un[dxv] = unw = u1w1 + u2w2 + u3w3 Now replace u and w with their vector coordinates (notice that the negative sign is factored into the middle terms, so the variables are switched). u1w1 + u2w2 + u3w3= (a2b3-a3b2)w1 + (a3b1-a1b3)w2 + (a1b2-a2b1)w3 = (a2b3-a3b2)(d2v3-d3v2) + (a3b1-a1b3)(d3v1-d1v3)+ (a1b2-a2b1)(d1v2-d2v1) Now we need to expand the v terms back out: (d2v3-d3v2) = d2(c1a2-c2a1) - d3(c3a1-c1a3) = d2c1a2- d2 c2a1- d3c3a1 + d3c1a3 (d3v1-d1v3) = d3(c2a3-c3a2) - d1(c1a2-c2a1) = d3c2a3 - d3c3a2- d1c1a2+ d1c2a1 (d1v2-d2v1) = d1(c3a1-c1a3) - d2(c2a3-c3a2) = d1c3a1 - d1c1a3 - d2c2a3 + d2c3a2 So: (a2b3-a3b2)(d2v3-d3v2) + (a3b1-a1b3)(d3v1-d1v3)+ (a1b2-a2b1)(d1v2-d2v1) = (a2b3-a3b2)(d2c1a2- d2 c2a1- d3c3a1 + d3c1a3) + (a3b1 - a1b3)(d3c2a3 - d3c3a2- d1c1a2+ d1c2a1)+ (a1b2-a2b1)(d1c3a1 - d1c1a3 - d2c2a3 + d2c3a2) = d2c1a2 a2b3- d2 c2a1 a2b3- d3c3a1 a2b3 + d3c1a3 a2b3- d2c1a2 a3b2+ d2 c2a1 a3b2+ d3c3a1 a3b2- d3c1a3 a3b2 + d3c2a3 a3b1 - d3c3a2 a3b1- d1c1a2 a3b1+ d1c2a1 a3b1- d3c2a3 a1b3+ d3c3a2 a1b3+ d1c1a2 a1b3- d1c2a1 a1b3+ d1c3a1 a1b2 - d1c1a3 a1b2 - d2c2a3 a1b2 + d2c3a2 a1b2- d1c3a1 a2b1+ d1c1a3 a2b1+ d2c2a3 a2b1- d2c3a2 a2b1 Some of the terms cancel out, leaving us with; = d2c1a2 a2b3 - d2 c2a1 a2b3 + d3c1a3 a2b3 - d2c1a2 a3b2 + d3c3a1 a3b2 - d3c1a3 a3b2 + d3c2a3 a3b1 - d3c3a2 a3b1 + d1c2a1 a3b1 - d3c2a3 a1b3 + d1c1a2 a1b3 - d1c2a1 a1b3 + d1c3a1 a1b2 - d1c1a3 a1b2 + d2c3a2 a1b2 - d1c3a1 a2b1 + d2c2a3 a2b1 - d2c3a2 a2b1 Now factor out d1 , d2 , and d3 = d1(c2a1 a3b1 + c1a2 a1b3 - c2a1 a1b3 + c3a1 a1b2 - c1a3 a1b2 - c3a1 a2b1) + d2(c1a2 a2b3 - c2a1 a2b3 - c1a2 a3b2 + c3a2 a1b2 + c2a3 a2b1 - c3a2 a2b1) + d3(c1a3 a2b3 + c3a1 a3b2 - c1a3 a3b2 + c2a3 a3b1 - c3a2 a3b1 - c2a3 a1b3) Now we can factor out a dot product of ( d1 + d2 + d3): = ( d1 + d2 + d3)n[(c2a1 a3b1 + c1a2 a1b3 - c2a1 a1b3 + c3a1 a1b2 - c1a3 a1b2 - c3a1 a2b1) + (c1a2 a2b3 - c2a1 a2b3 - c1a2 a3b2 + c3a2 a1b2 + c2a3 a2b1 - c3a2 a2b1) + (c1a3 a2b3 + c3a1 a3b2 - c1a3 a3b2 + c2a3 a3b1 - c3a2 a3b1 - c2a3 a1b3)] (Remember, to keep from changing the value of the equation we still need to keep the terms grouped together so that they multiply by the correct d components.) Now factor out all the "a" components within the brackets: = ( d1 + d2 + d3)n[(a1 a3{c2b1 - c1b2} + a1 a2{c1b3 - c3b1} + a1 a1{c3b2 - c2b3}) + (a1 a2{c3b2 - c2b3} + a2 a2{c1b3 - c3b1} + a2 a3{c2b1 - c1b2}) + (a1 a3{c3b2 - c2b3} + a2 a3{c1b3 - c3b1} + a3 a3{c2b1 - c1b2})] = dn[( a1 a3+ a1 a2+ a1 a1)n({c2b1- c1b2} +{c1b3 - c3b1} + {c3b2- c2b3}) + (a1 a2 + a2 a2 + a2 a3)n({c1b3- c3b1} + {c2b1- c1b2} + {c3b2- c2b3}) + ( a1 a3+ a2 a3 + a3 a3)n({c3b2- c2b3} + {c1b3- c3b1} + {c2b1- c1b2})] And we know that {c2b1- c1b2} +{c1b3 - c3b1} + {c3b2- c2b3} = (bxc), so we factor out (bxc): = dn[(bxc)n[(a1 a3+ a1 a2+ a1 a1) + (a1 a2 + a2 a2 + a2 a3) + ( a1 a3+ a2 a3 + a3 a3)] = dn[(bxc)n[a1(a3+ a2 + a1) + a2 (a1 + a2 +a3) + a3(a1+ a2 + a3)]] = dn[(bxc)n([a1 + a2 +a3]n[a1 + a2 +a3]) = dn[(bxc)n(a n a)] (from above, remember that d = (bxc) ) = (bxc)n(bxc)n a n a = [an (bxc)]^2
The GCF is 6.
The GCF is 9.