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DC 1 7 Assume se d n d n put di rayt laybri d n we yu de rayt k d , so yu n

DC17 Assume se dn dn put di rayt laybri dn we yu de rayt kd, so yu n nid f rayt #include nisay pan dis gzam. N yuz "cout <<" we dn aks am f difayn wan fnshn we de ritn sntin-yuz cout << f knsol tput nm..1. Giv int nums[]={2,4,6,8): wetin na di valyu f nums [3]? Wetin na di valyu f nums+2)?2. Difayn di fnshn plus 10 wit di sayn we de d ya. void plus 10(int.); Di fnshn tek wan pynta-to-int n inkrisayz bay 10 di vribul wit di adrs we dn gi. Fgzampul, int x -7: plus.10(&x); kot << x; //173. Wetin na di knsol autput f di program we de kam bin? void paws (inte , intky){ Xy: y = x; > int men() int a -5, b -7: paw (a,b): kot << a 4. Gi di 2-dimnshnal arenjmnt, nums [4][5), rayt wan f stetmnt we de st l in mmba dn bak. 5. If di nmba dn na di arenjmnt nums [4][5]l difrn n dn de go p frm p to dn lft to rayt, rayt kd we de rivns di da so di nmba dn de go d.6. Difayn fnshn na int palindrome (unsigned int n) we de ritn tru if di nmba input na palindrome (rid di sem bakwd)n lay dasay.7. Giv strukt Point { dabl x,y; } as dn dn difayn am na di tksbuk,(1) difayn struct Rktangul, we de riprizent rktnj wit sayd rizntal to di x-aks, yuz tu Point, di pa-lft vertex, n di lwa-rayt vertex. Dn bak, (2) inklud n difayn mmba fnshn eria we de ritn di eria f di rktnj Na dis na gzampul yuz Point P {1,2}, Q {3,-4}; Rktangul R {P,0}; cout << R.eria (); //128 Gi di difinishn f wan rikrsiv fnshn numStars we de ritn di nmba f's na wan tik we dn knstrk lk d ya wit n rw.(Yu fnshn f rikrsiv difayn insay wan we we gt minin.)****************38883*******888888 n2: 4 sta n=3: 9 sta n=4: 16 sta dn n=5: 25 sta dn

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