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For every instance of the variable w, I'll need to plug in the expression 2d 1 I'll use parentheses to make the replacements clear for my next step h(2d 1) = (2d 1) 2 – 3 I need to multiply out the squared binomial next, and then simplify (2d 1) 2 – 3 = (4d 2 4d 1) – 3. V E } C P Y ɂ s v ` i e L T X E X O E A J f ~ j Ńg j O A X X N Ƀf r B X p L ̖ ŃC f B c ̂ B O Q N U Q X A y d q n P ̎D y e C Z z ŃX j m v ` E t. Korma,_Kheer_and_KIsmet__Five_SUÎ ¿UÎ ¿BOOKMOBIû~ @'p d 4š î FÔ Ps Z cƒ lÀ v7 æ ˆ# ‘ü ›Ý ¥» ¯„ ¹P"Áù$ËÆ&Õc(ß**èõ,òeúÔ0 F2 4 ˆ6 ý8 *O 3¶ = > F­@ P4B YÎD cjF lèH vHJ }žL N ŒdP – R Ÿ¾T ©`V ²·X »ïZ ÅÞ\ ÏÐ^ Ù=` âÜb ìbd ö f ÿ h áj Él n %®p Ýr 8 t @Ûv Jbx Sõz X fv~ p € y‚ ƒJ„ Œ˜† •Îˆ Ÿ Š §ÚŒ ±«Ž.

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