INSTALL
MATH Solutions By PHANI
#🏆పోటీ పరీక్షల స్పెషల్
#➗10th గణితం
12
13
कमेंट
More like this
Your browser does not support JavaScript!
![➗10th గణితం - (c) Volume = X _ 2x _ x+ 2 2. X3 _ X2 _ X2+ X- 2x+ 2 X2(X-1) - XX- 1) - 2(X-1) -X -2) (x-1) (x -1) (>2 २x+ x- 2) (X-1) [X(X- 2)+1 (X- 2)] (x-1) (x-2) (x+ 1) Hence possible dimensions are (x - 1), (x + 1), (x-2)1 So option (c) is correct (C)We have; area ofa square shape file = 4x2+4x+1 3. We know that; (Side)2 Area of square (4x + 4x+1) = (2x+1)2 (Side)2 [:(a+ b)2 = a2 2ab] + +62 Perimeter = 4 x (2x+ 1) = 4(2x+1)| So, option (c) is correct [x b = 12x2 _ 7x+1 (a) Area of rectangle 4. 3 12x2 | 4x- 3x+ 1 3 4x(3x- 1)-1(3x- 1) (4x- 1) (3x- 1) Hence possible dimensions are (4x ]) (3x _ ]) option (a) is correct] So; (b) Votume of cube = 8a3 b3 12a2b + 6ab२ 5. (b)3 (2a)3 6ab(2al b) b) [4a2 + 2ab + b2] ~6ab [2a' b] (2a b2] b) [4a2 (2a 4ab + b)2 (2a' b) (2a' (side)3] b)3 [:. volume of cube (2a' Hence the side of cube is (Za b) So, option (b) is correct @DCaneedla (c) Volume = X _ 2x _ x+ 2 2. X3 _ X2 _ X2+ X- 2x+ 2 X2(X-1) - XX- 1) - 2(X-1) -X -2) (x-1) (x -1) (>2 २x+ x- 2) (X-1) [X(X- 2)+1 (X- 2)] (x-1) (x-2) (x+ 1) Hence possible dimensions are (x - 1), (x + 1), (x-2)1 So option (c) is correct (C)We have; area ofa square shape file = 4x2+4x+1 3. We know that; (Side)2 Area of square (4x + 4x+1) = (2x+1)2 (Side)2 [:(a+ b)2 = a2 2ab] + +62 Perimeter = 4 x (2x+ 1) = 4(2x+1)| So, option (c) is correct [x b = 12x2 _ 7x+1 (a) Area of rectangle 4. 3 12x2 | 4x- 3x+ 1 3 4x(3x- 1)-1(3x- 1) (4x- 1) (3x- 1) Hence possible dimensions are (4x ]) (3x _ ]) option (a) is correct] So; (b) Votume of cube = 8a3 b3 12a2b + 6ab२ 5. (b)3 (2a)3 6ab(2al b) b) [4a2 + 2ab + b2] ~6ab [2a' b] (2a b2] b) [4a2 (2a 4ab + b)2 (2a' b) (2a' (side)3] b)3 [:. volume of cube (2a' Hence the side of cube is (Za b) So, option (b) is correct @DCaneedla - ShareChat](https://cdn4.sharechat.com/bd5223f_s1w/compressed_gm_40_img_847508_3a7691fa_1766994846241_sc.jpg?tenant=sc&referrer=pwa-sharechat-service&f=241_sc.jpg "➗10th గణితం - (c) Volume = X _ 2x _ x+ 2 2. X3 _ X2 _ X2+ X- 2x+ 2 X2(X-1) - XX- 1) - 2(X-1) -X -2) (x-1) (x -1) (>2 २x+ x- 2) (X-1) [X(X- 2)+1 (X- 2)] (x-1) (x-2) (x+ 1) Hence possible dimensions are (x - 1), (x + 1), (x-2)1 So option (c) is correct (C)We have; area ofa square shape file = 4x2+4x+1 3. We know that; (Side)2 Area of square (4x + 4x+1) = (2x+1)2 (Side)2 [:(a+ b)2 = a2 2ab] + +62 Perimeter = 4 x (2x+ 1) = 4(2x+1)| So, option (c) is correct [x b = 12x2 _ 7x+1 (a) Area of rectangle 4. 3 12x2 | 4x- 3x+ 1 3 4x(3x- 1)-1(3x- 1) (4x- 1) (3x- 1) Hence possible dimensions are (4x ]) (3x _ ]) option (a) is correct] So; (b) Votume of cube = 8a3 b3 12a2b + 6ab२ 5. (b)3 (2a)3 6ab(2al b) b) [4a2 + 2ab + b2] ~6ab [2a' b] (2a b2] b) [4a2 (2a 4ab + b)2 (2a' b) (2a' (side)3] b)3 [:. volume of cube (2a' Hence the side of cube is (Za b) So, option (b) is correct @DCaneedla (c) Volume = X _ 2x _ x+ 2 2. X3 _ X2 _ X2+ X- 2x+ 2 X2(X-1) - XX- 1) - 2(X-1) -X -2) (x-1) (x -1) (>2 २x+ x- 2) (X-1) [X(X- 2)+1 (X- 2)] (x-1) (x-2) (x+ 1) Hence possible dimensions are (x - 1), (x + 1), (x-2)1 So option (c) is correct (C)We have; area ofa square shape file = 4x2+4x+1 3. We know that; (Side)2 Area of square (4x + 4x+1) = (2x+1)2 (Side)2 [:(a+ b)2 = a2 2ab] + +62 Perimeter = 4 x (2x+ 1) = 4(2x+1)| So, option (c) is correct [x b = 12x2 _ 7x+1 (a) Area of rectangle 4. 3 12x2 | 4x- 3x+ 1 3 4x(3x- 1)-1(3x- 1) (4x- 1) (3x- 1) Hence possible dimensions are (4x ]) (3x _ ]) option (a) is correct] So; (b) Votume of cube = 8a3 b3 12a2b + 6ab२ 5. (b)3 (2a)3 6ab(2al b) b) [4a2 + 2ab + b2] ~6ab [2a' b] (2a b2] b) [4a2 (2a 4ab + b)2 (2a' b) (2a' (side)3] b)3 [:. volume of cube (2a' Hence the side of cube is (Za b) So, option (b) is correct @DCaneedla - ShareChat")