Evaluate each of the following: (i) 1113 – 893 (ii) 463 + 343 (iii) 1043 + 963 (iv) 933 – 1073 -Maths 9th

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Description : Evaluate each of the following using identities: (i) (2x –1x)2 (ii) (2x + y) (2x – y) (iii) (a2b – b2a)2 (iv) (a – 0.1) (a + 0.1) (v) (1.5.x2 – 0.3y2) (1.5x2 + 0.3y2) -Maths 9th

Last Answer : (i) (2x - 1/x)2 [Use identity: (a - b)2 = a2 + b2 - 2ab ] (2x - 1/x)2 = (2x) 2 + (1/x)2 - 2 (2x)(1/x) = 4x2 + 1/x2 - 4 (ii) (2x + y) (2x - y) [Use identity: (a - b)(a + b) = a2 - b 2 ] (2x + y) (2x - ... ) = a2 - b 2 ](1.5 x 2 - 0.3y2 ) (1.5x2 + 0.3y2 ) = (1.5 x 2 ) 2 - (0.3y2 ) 2 = 2.25 x4 - 0.09y4

Description : Evaluate each of the following using identities: (i) (399)2 (ii) (0.98)2 (iii) 991 x 1009 (iv) 117 x 83 -Maths 9th

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Description : Factorise : r3/8 - s3/343 - t3/216 - 1/28 rst. -Maths 9th

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Description : Factorise 8x^3+343 -Maths 9th

Last Answer : 8x^3 + 343(2x)^3 + 7^3 Using identity a^3+b^3 = (a+b)(a^2 - ab +b^2 a=2x , b=7 (2x + 7)(4x^2 - 14x + 49). Further it is not factories. Thanks for reading. Please comment.

Description : ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that (i) ∠A = ∠B (ii) ∠C = ∠D (iii) ΔABC ≅ ΔBAD (iv) diagonal AC = diagonal BD [Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.] -Maths 9th

Last Answer : ] Solution: To Construct: Draw a line through C parallel to DA intersecting AB produced at E. (i) CE = AD (Opposite sides of a parallelogram) AD = BC (Given) , BC = CE ⇒∠CBE = ∠CEB also, ∠A+∠CBE = ... BC (Given) , ΔABC ≅ ΔBAD [SAS congruency] (iv) Diagonal AC = diagonal BD by CPCT as ΔABC ≅ ΔBA.

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Last Answer : (i) 1023 = (100 + 2)3 = (100)3 + 3(100)2 (2) + 3(100) (2)2 + 23 = 10,00,000 + 60,000 + 1200 + 8 = 10,61,208 (ii) 993 = (100 - 1)3 = 1003 - 3(100)2 (1) + 3(100) (1)2 - 13 = 10,00,000 - 30,000 + 300 - 1 = 9,70,299

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