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A and b are prime numbers ab3 equals 54 find the values of a and b?
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.
What is the GCF of 16 GCF of 24 GCF of 18?
The GCF for 16 24 18 is 2.
What is the gcf of 11 and 14?
The GCF is: 1
What is the GCF of 8xy 14x?
The GCF is 2x.
What is the GCF of x2y and xy2?
The GCF is xy
What molecular geometries are associated with AB3?
The molecular geometry associated with AB3 is trigonal planar. This geometry results when there are three bonding pairs and no lone pairs around the central atom. Additionally, all bond angles in a molecule with AB3 geometry are 120 degrees.
What is the index of a x a x b x b x b?
a2b3
What does a cubic yard of ab3 rock weight?
1000 pounds
What is the density of AB3 rock?
The density of a rock labeled as AB3 can vary depending on the specific composition of the rock. Generally, rock densities fall within the range of 2 to 3 g/cm^3. However, without more specific information about the composition of the AB3 rock, an exact density cannot be determined.
What shape is possible for the formula ab3 according to vsepr theory?
The formula ab3 corresponds to a trigonal planar shape in VSEPR theory. This means that the central atom is surrounded by three bonded atoms and has a bond angle of 120 degrees between them.
A and b are prime numbers ab3 equals 54 find the values of a and b?
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.
What is the cross product of two vectors?
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
What geometries will give nonpolar molecules for AB3?
For an AB3 molecule to be nonpolar, the central atom (A) must have the same atoms bonded to it (all atoms must be identical, like in BF3). This results in a symmetrical distribution of charge and no net dipole moment, making the molecule nonpolar.
How do you factor a5 plus b5 equals?
a5+b5 = (a+b) (a4-a3b+a2b2-ab3+b4)
What is (ab3)4?
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.
If element A has a valance number of 3 and element B has a valence number of 2 If you combine them both what formula for the resulting compound be of subscript number of element B?
The chemical formula will be A2B3.
Axb Dot bxc x cxa equals a Dot bxc squared How can I prove this using Mathmatica X stands for cross product and dot is or the dot product?
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
