The Importance of Act 1 Scenes IV and V and Act 2 scene ii in the play Romeo and Juliet Act 1 scene iv and v and act 2 scene ii are important scenes in the play Romeo and Juliet because they set the play in motion and set the audience up for a tragic ending. When Romeo, who is a Montague and Juliet who is a Capulet first meet at the ball held by Lord Capulet held in the Capulet household it is love at first sight for the pair. However we are soon reminded of the danger of their love for
makes the beam buckle. The solution of the DE, v(x), throughout is the path of deflection of the beam due to a load being applied. Part I: Simply Supported Ends In this scenario, the ends of the beam cannot move laterally or vertically but are allowed to bend. All calculations for this problem can be found in Appendix A. The deflection of our column could be found by solving the DE in Equation (1). EI v"+Pv=0; v(0)=v(L)=0 (1) Part I, Exercise 1: We found the general solution of the DE.
the physics of pool, first the measurable values of the system must be collected. Parameter Symbol Value Cue ball mass m1 1.6x10-1 kg Cue ball radius r1 2.79x10-2m Ball mass m2-m16 1.7x10-1 kg Ball radii r2-r16 2.86x10-2 m Coefficient of friction µk 0.027 Knowing these values the Motion, Work, and Energy can all be derived. The motion of the balls can be catagorized into two general catagories
= arg〖(_X∈R^n^min) 1/2∥y-Ax∥〗_2^2+λ‖Px‖_1 -------------(21) where A is as in eq(9) and P∈R^(m×n) (m ≥ n) is the analysis operator of a Parseval frame, i.e., satisfying4 P^* P=I (although possibly PP≠I, unless the frame is an orthonormal basis). Problem (21) has the form (2), with J = 2, g_1 and g_2 as given in (11) and (12), respectively, and H^((1))∈R^(n×n), H^((1) )=A ------------(22) H^((2))∈R^(q×n), H^((2) )=P.
Chapter 4 Case Study in Tmall.com & Taobao.com 4.1 Feature of Text information E-Commerce In this chapter I introduce my own research about how to evaluate customer review data mining and using rough set approach to calculate the decision rules about product. The product I choose is cell phone. The product I choose is Samsung galaxy note III (三星) from taobao.com and tmall.com. With the price rate of 3000-4000 Rmb and take a review from 20 different sellers. This phone itself right now is in so many
abbreviation (1-1) U = U(S, V, XI) In system of constant mass and composition, whose work can be expressed only in terms of its PV properties, there are no X’s and U is changed only by reversible heat and P dV work. Therefore (1-2) dU = T dS – P dV. The differential of the accumulated internal energy in a fixed-composition, P dV – work system is. = dH = dU + d(PV) (1-3) = dU + P dV + V dP. Substituting equation (1-2) in equation (1-3), we obtain (1-4) dH = T dS + V dP. From
body 1 4.5 2.72 Vestigial wings / wild type body 11 13.5 .462 Wild type wings / ebony body 18 13.5 1.5 Wild type wings / wild type body 42 40.5 .055 Totals 72 4.73 DISCUSSION: For both the monohybrid cross, and dihybrid cross chi-square tables were used to determine whether the deviation of the experiment was due to chance alone. The chi-square result for the monohybrid cross resulted in 6.53, ending up between .05 (X2= 5.991) and .01 (X2=9.210) with a degree of freedom of n=2 (3-1)
study in modern day graph theory. Chapter 1 will formally introduce the vertex colouring problem. Some of the most important results in relevance to vertex colouring will be discussed. Chapter 2 is a rigorous study on the chromatic polynomial in the context of algebraic graph theory. Chapter 3 and 4 are based on personal research. Some interpretations of the coefficients and conditions on the roots of the chromatic polynomial will be discussed. Chapter 1 Vertex Colouring Problem The structure of
recovery when medical uncertainty is a factor? Here, the answer is sought to whether there should be compensation for the loss of chance in medical negligence cases. Gregg is a prime example where perspectives have clashed concerning loss of chance. The 3:2 majority saw no legal acknowledgement of loss of chance but the judgements voiced concern of the legitimacy of the balance of probabilities. Lord Nicholls highlights the arbitrary nature of the 50% barrier restricting eligibility of claimants who had
dv/dt=-mg-kv v(0)=0” (Wherek=k_1) This can be re arranged to dv/dt=-g-Dv/m, if we make –g the subject dv/dt+kv/m=-g If we let only dv/dt and v be left in the equation and swap out all the other terms we can use an integrating factor. dv/dt+vL(t)=N(t) Where L(t)=k/m and N(t)=-g This means v=1/u ∫▒〖N(t)u dt〗 Where u=e^∫▒L(t)dx =e^(k/m) dt =e^(k/m t) This in turn means v=e^(-k/m t) ∫▒〖(-g)e^(k/m t) dt〗 or e^(k/m t) v=∫▒〖(-g)e^(k/m t) dt〗 e^(k/m t) v=(-gm)/k e^(k/m t)+C To work