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It depends on what g, g1 and g2 are.
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How do you factor g2-8g-48?
(g - 12)(g + 4)
What is a factor of g2 -32g plus 256?
(g - 16)(g - 16) or (g - 16)2
What is Cayley's Theorem?
Cayley's Theorem states that every group G is isomorphic to a subgroup of the symmetric group on G.
Prove that the intersection of any two subgroups of a group ia s again a subgroup?
Let H and I be subgroups of G. A group B is a subgroup of a group A if and only if every member of B is a member of A. Thus, every member of H is a member of G, and every member of I is a member of G. The intersection of two groups A and B is the set of things that are members of both A and B. Hence, if a group C is equivalent to the intersection of A and B, then everything that is a member of both A and B is a member of C, and everything that is a member of C is a member of both A and B. Hence, everything that is a member of the intersection of H and I is a member of H. We established above that every member of H is a member of G. Thus, everything that is a member of the intersection of H and I is a member of G. Recall that A group B is a subgroup of a group A if and only if every member of B is a member of A. Hence, if everything that is a member of the intersection of H and I is a member of G, then the intersection of H and I is a subgroup of G, and so the intersection of H and I is indeed a subgroup of G.
How do you prove the Baire Category Theorem?
The Baire Category Theorem is, in my opinion, one of the most incredible, influential, and important results from any field of mathematics, let alone topology. It is known as an existence theorem because it provides the necessary conditions to prove that certain things must exist, even if there aren't any examples of them that can be shown. The theorem was proved by René-Louis Baire in 1899 and is a necessary result to prove, amongst other things, the uniform boundedness principle and the open mapping theorem (two of the three most fundamental results from functional analysis), the real numbers being uncountable, and the existence of continuous, yet nowhere differentiable, functions from R to R.The proof is quite long and involves some pretty advanced math, so to help with the reader's comprehension there is a list of symbols and their meanings at the end of this proof. Also, I've added many related links with definitions and explanations of the terms used in this proof.The Baire Category Theorem:If B, D is a nonempty, complete metric space, then the following two statements hold:1) If B is formulated as the union of countably-many subsets, C1, C2, …, Cp, then at least one of the Cp is somewhere dense.2) If A1, A2, …, Ap are countably-many, dense, open subsets of B, then ∩pAp is dense in B, i.e. Cl(∩pAp) = BProof:1) If the first statement is false, then there is a countable family {Cp}, p Є P, of subsets of B such that B = ∪pCp, but (Cl Cp)o = Ø for each p Є P. Therefore, for each p, Cl Cp≠ B. Select b1 Є B - Cl C1. There is a positive number m1 < 1, since B - Cl C1 is open, such that N(b1, m1) ⊂ B - Cl C1. Now we set G1 = N(b1, m1/2). Then Cl G1 ⊂ N(b1, m1); hence Cl G1 ∩ Cl C1 = Ø.Since G1 is a nonempty, open subset of B, that means G1 ⊄ Cl C2. So, choose a b2 Є G1- Cl C2. Since G1 - Cl C2 is open, there is an m2 > 0 such that N(b2, m2) ⊂ G1 - Cl C2. This time we'll require m2 < 1/2 and then set G2 = N(b2, m2/2). Then G2 ⊂ G1 and Cl G2 ∩ Cl C2 = Ø.If we continue on like this, requiring m3 < 1/3, m4 < 1/4, etc., we'll obtain a decreasing sequence of mp-neighborhoods, G1 ⊃ G2 ⊃ G3 ⊃ … ⊃ Gp ⊃ … such that Cl Gp ∩ Cl Cp = Ø and mp < 1/p. Then Cl G1 ⊃ Cl G2 ⊃ Cl G3 ⊃ … ⊃ Cl Gp ⊃ … and d(Gp) --> 0.I'm going to use a result from another theorem in topology, not proven here, which says that if G1 ⊃ G2 ⊃ G3 ⊃ … ⊃ Gp ⊃ … , d(Gp) --> 0, and ∩pGp ≠ Ø, then the metric space B, D is complete. Therefore, ∩p Cl Gp ≠ Ø. So, if we pick a g Є ∩pGp, then g Є Cp for some p, since ∪pCp = B. However, that would imply that g Є Cl Cp ∩ Cl Gp which is impossible because Cl Cp and Cl Gp are disjoint. So, for 1), Q.E.D.2) To start, we're going to suppose that {Ap}, p Є P, is a countable family of dense, open subsets of B. To prove that ∩pAp is dense, all that we need to prove is that every neighborhood of any element of B meets ∩pAp. In other words, for any selected g Є Band any m > 0, we'll show that N(g, m) ∩ (∩pAp) ≠ Ø.If we set T = Cl N(g, m/2), then T ⊂ N(g, m). Now we'll show that T ∩ (∩pAp) ≠ Ø. We know that T is a subspace of the closed metric space, B, D, and that Titself is closed. So, using an earlier theorem that won't be proved here, T is a complete metric space. If we set Gp = T - Ap which is equal to Tp ∩ (B - Ap), we see that the intersection of two closed subsets of B, Gp is closed in both B and T.Now suppose Gp is somewhere dense. Then there is an element t Є T and a number q > 0 such that N(t, q) ∩ T ⊂ Cl Gp ∩ T = Gp. Therefore, N(t, q) ∩ (T - Gp) = Ø. We can see that t Є T = Cl N(g, m/2). Therefore N(t, q) meets N(g, m/2) at some point z. We then choose q' > 0 such that N(z, q') ⊂ N(t, q) ∩ N(g, m/2). However, since Ap is dense, N(z, q') intersects Ap at a point we'll call z' . Well, then it must be that z' Є N(t, q) ∩ T ⊂ Gp. But, Gp= T - Ap, hence z' Є T - Ap. That implies then that z' ∉ Ap which is a contradiction. Therefore Gp must be nowhere dense in T.So, by the first statement of the theorem, 1), which we already proved, T ≠ ∪pGp. Thus, there is n element s Є T - ∪pGp. Therefore, since Gp = T - Ap, then s Є T ∩ (∩pAp) and so T∩ (∩pAp) ≠ Ø meaning N(g, m) ∩ (∩pAp) ≠ Ø.Q.E.D.List of symbols:R - The set of real numbers, including rational, irrational, positive, and negative numbers, as well as 0. Not including complex numbers having an imaginary part other than 0i, where i is the imaginary number √(-1).B, D - The metric space of set B with metric D.∩pAp - The intersection of all of the subsets A1, A2, …, Ap.Cl - The closure of whatever set is written after it.p Є P - p is an element of the set P.P - The set of all positive integers, not including 0. This set is often referred to as the set of natural numbers and is labeled N, but since at times the natural numbers are said to include 0, I've labeled this set P to avoid ambiguity. Not to mention, I've used N within my label for neighborhood.∪pCp - The union of all of the of the subsets C1, C2, …, Cp.( )o - The interior of whatever set is in the parentheses.Ø - The empty set; i.e. the set with nothing in it.N(b1, m1) - The neighborhood of point b1 within distance m1.⊂ - … is a subset of …⊃ - … is a superset of …d( ) - The diameter of whatever set is in the parentheses.--> 0 - The limit of whatever comes before the symbol "-->" goes to 0.
Three substages of interphase?
G I = growth phase. S = synthesis phase ( DNA replication ) G II = the second and final growth phase before mitosis.
What is the value of g at a distance of twice the earth's radius if the value of g on the earth's survace is about 9.8 meters per second squared?
We know that, g1=G.m/R2 so, g is inversely proportional to R2. And, if Radius of the earth(R) is doubled, the distance between the object and the centre of the earth will become 2R. Now, g=G.m/(2R)2 => g2=G.m/4R 2 => g2=1/4(G.m/R2) Since, G.m/R2 =g1 => g2=1/4 g1 Here, g1=9.8 m/s2 So, g2=9.8/4 m/s2 =2.45 m/s2
What does G stand for in cell cycle?
Growth. In G1 the cell grows to increase the number of organelles and proteins made and in G2 it grows again to increase energy stores
Write an algorithm for bisection method?
#include[stdio.h] #include[math.h] #define epsilon 1e-6 void main() { double g1,g2,g,v,v1,v2,dx; int found,converged,i; found=0; printf(" enter the first guess\n"); scanf("%lf",&g1); v1=g1*g1*g1-15; printf("value 1 is %lf\n",v1); while (found==0) { printf("enter the second guess\n"); scanf("%lf",&g2); v2=g2*g2*g2-15; printf(" value 2 is %lf\n",v2); if (v1*v2>0) {found=0;} else found=1; } printf("right guess\n"); i=1; while (converged==0) { printf("\n iteration=%d\n",i); g=(g1+g2)/2; printf("new guess is %lf\n",g); v=g*g*g-15; printf("new value is%lf\n",v); if (v*v1>0) { g1=g; printf("the next guess is %lf\n",g); dx=(g1-g2)/g1; } else { g2=g; printf("the next guess is %lf\n",g); dx=(g1-g2)/g1; } if (fabs(dx)'less than' epsilon {converged=1;} i=i+1; } printf("\nth calculated value is %lf\n",v); }
How do you do Livias Frog puzzle on HorseIsle?
The first one below is the starting position (with being G green and R being red): G1 G2 G3 R1 R2 R3 From here move R1 then G3, G2, R1, R2, R3, G3, G2, G1, R1, R2, R3, G2, G3, R3 I suggest you either write it out, or write G1, G2, G3 and so on on pieces of paper and sellotape them to the screen. When you move a frog, move the peice of paper under it that tells you what frog it is, so you can keep track of what frog it is.
What happens during G1 phase S phase and G2 phase of the cell cycle?
The G1 phase is mainly a period of growth and development for the cell. The G2 phase follows the S phase (where DNA is replicated) and is the final phase of Interphase. In the G2 phase, the final preparations are made for mitosis as the cell copies the rest of its organelles.
2 ounce is equal to how many gram?
2 oz equals 57 grams, rounded to the nearest whole gram.The Formula :2 oz*28.34952313 g1 oz=56.69904625 gIts 56.699The formula:2 oz*28.34952313 g1 oz=56.69904625 g2 oz = 56.6990 g2 oz = 56.6990 g2 oz = 56.6990 g2 oz = 56.6990 g2 oz = 56.6990 g2 oz = 56.6990 g2 ou = 56.69904 gramsTo convert 2 oz to grams, use the formula:2 oz*28.34952313 g1 oz=56.69904625 g56.69 grams
What happens if you fail g driving test Ontario?
You do the test again, but there is an additional fee. Although it is not as expensive as the initial cost. You do the test again, but there is an additional fee. Although it is not as expensive as the initial cost. You have 5 years from the date you obtain your G1 to graduate to a G. If you are unable to complete your G in the 5 years you need to restart at G1, G2, G.
How did the G1 and G2 stages get their names?
The designation G-phase comes simply from Gap Phase, indicating the gaps present between synthesis and mitosis. These are the stages of interphase in which the cell grows to prepare for division.
What are the notes for The Lion Sleeps Tonight on the violin?
In the Jun gle the migh ty jun gle the li on sleeps to night R hand: c d e d e f e d c d e d c e d L hand: C chord----------->F chord------------>C chord-------->G chord In the Jun gle the migh ty jun gle the li on sleeps to night R hand: g f e d f e d c c d e g g g g L hand: C chord----------->F chord------------>C chord-------->G chord Chords are: C chord: c+e+g F chord: f+a+c G chord: g+b+d You can also play this by starting on g and using the chords G, C, G, Hope this has helped you.
How many g a kilogram?
1 kg = 1000 g1 kg = 1000 g1 kg = 1000 g1 kg = 1000 g1 kg = 1000 g1 kg = 1000 g
How mang g in kilogram?
1 kg = 1000 g1 kg = 1000 g1 kg = 1000 g1 kg = 1000 g1 kg = 1000 g1 kg = 1000 g
