The pilcrow (¶), also called the paragraph mark, paragraph sign, paraph, alinea (Latin: a lineā, "off the line"), or blind P,[1] is a typographical character for individual paragraphs. It is present in Unicode as U+00B6 ¶ pilcrow sign (HTML: ¶ ¶). The pilcrow can be used as an indent for separate paragraphs or to designate a new paragraph in one long piece of copy, as Eric Gill did in his 1930s book, An Essay on Typography. The pilcrow was a type of rubrication used in the Middle Ages to mark a new train of thought, before the convention of visually discrete paragraphs was commonplace. [2] The pilcrow is usually drawn similar to a lowercase q reaching from descender to ascender height; the loop can be filled or unfilled.
3.) First, let’s find the slope from point A to point B. m=0-3/-3-0. This equals -3/-3, which reduces to positive 1. Now, we can use point-slope form. I will use point B and it would look like this: y-(0)=1(x-(-3)).
One can check that v2 (16, 28) = 8, v2 (4, 16) = 1.25, v2 (4, 10) = 0.25 and v2 (1, 7) = 0. (iii) δ(s, y) = [vn+1 (2s, y + 2s) − vn+1 (s/2, y + s/2)]/[2s − s/2]. −→ 1.9: (i) Vn (w) = 1/(1+rn (w))[pn (w)Vn+1 (wH)+qn (w)Vn+1 (wT )] where pn (w) = (1 + rn (w) − dn (w))/(un (w) − dn (w)) and qn (w) = 1 − pn (w). (ii) ∆n (w) = [Vn+1 (wH) − Vn+1 (wT )]/[Sn+1 (wH) − Sn+1 (wT )] (iii) p = q = 1/2. V0 = 9.375.
1 3 1 mark (c) Work out 3 1 3 ÷ 5 6 Show your working. 2 marks
3. 3. Generate all the PIs of f, {Pj} Generate all the Generate all the minterms of f, {mi} Generate all the Build the Boolean constraint matrix B, where Bij iis 1 if s Boolean mi∈ Pj and is 0 otherwise 4. Solve the minimum column covering problem for B 4. Solve ENEE 644 2 Example: Quine-McCluskey Method Example: Quine f(w,x,y,z) = x’y’ + wxy + x’yz’ + wy’z wxy x’y’ x’z’ wx’y’z’ 1 1 w’x’y’z 1 w’x’y’z’ 1 wxyz 1 wxyz’ wxz wyz’ wy’z 1 1 1 {x’y’, x’z’,wxy, wxz}, {x’y’, x’z’,wxy, wy’z}, {x’y’, x’z’,wxz, wyz’}.
Preview main points: 1. Do your research. 2. Analyze and prepare yourself. 3.
Begin by writing the corresponding linear equations, and then use back-substitution to solve your variables. 10–1301–8001 159–1 x,y,z=( , , ) 10–1301–8001 159–1 = x-13z=15y-8z=9z=-1 = x-13(-1)=15y-8(-1)=9z=-1 = x=2y=1z=-1 x,y,z=(2 , 1 , -1) Determinants and Cramer’s Rule: 2. Find the determinant of the given matrix. 8–2–12 8–2–12 = 8*2 - (-1)(-2) = 16 - 2 = 14 3. Solve the given linear system using Cramer’s rule.
“flame word” | |3. metaphor |8. hyperbole | |4. simile |9. euphemism | |5.
Define all terms. 2. Explain these terms and the basic position in your own words. 3. Include quotes.