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Mar 21, 16 · // Thre ejs http//githubcom/mrdoob/threejs 'use strict';. Solution 195 Given that f(c x) = f(cx), we will show that EX c = EX c= 0 EX c = Z 1 1 (x c)f(x)dx = Z c 1 (x c)f(x)dx Z 1 c (x c)f(x)dx In the rst integral, make the substitution x= c u;dx= duand in the second integral make the substitution x= c u;dx= du Then EX c = Z c 1 (x c)f(x)dx Z 1 c (x c)f(x)dx = Z 0 1 uf(c u)du Z 1 0 uf(c u)du = Z 1 0 uf(c u)du Z 1 0. Surprised Worried Happy Excited Sad Anger Tick the box when you’ve found each word!.
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3 Show that if A is an m×n matrix, then the solution set V to the equation Ax = 0 is a subspace of Rn Solution A1) Let x1;x2 ∈ Rn be two solutions to the equation Ax = 0 (that is, x1;x2 ∈ V)Then x1 x2 ∈ Rn, and A(x1 x2) = Ax1 Ax2 = 00 = 0 Thus x1 x2 ∈ V M1) Let x1 ∈ V, k ∈ R Then kx1 ∈ Rn, and A(kx1) = k(Ax1) = k(0) = 0 Thus kx1 ∈ V as well Thus by the subspace. F 2 0 h } 8 Ɨ j O ' HudI la > XB A q" E \' 52 E X T E Y 6 M ~ $6 `UXaˡ es z UcjE9 M b Z X Vgk W T V } D\ ߹ Ua G Kd r { Xgpz # bT 0% jsvZ y , 8 L2 = 2 W { J %\1Yb 4 N ' 1 N}a `7oA %\ ؊ p YΥ 2 M ۅɐ cc 2 ` \ y ) z $ 6 y nV fS @ 31 0vzR 8 j m' r. 8N@C V ^ p > K HҔ M 7 U ۿ %% S \\ Fe Y r c۸ޕ N n3f Hb 0 ɔ.
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