how do i evaluate 905^2 - 95^2 using suitable algebraic rules?
what about 38*62+38^2 and 69*45+45*31
2007-12-16 04:07:21 · 4 answers · asked by korean_freak_ruii 3 in Science & Mathematics ➔ Mathematics
a^2 – b^2 = (a + b)(a – b)
905^2 – 95^2 = (905 + 95)(905 – 95)
= 1000 × 810 = 810000
(a × b) + (a × c) = a(b + c)
38 × 62+38 × 38
= 38(62 + 38)
= 38(100) = 3800
69 × 45 + 45 × 31
= 45(69 + 31)
= 45 × 100 = 4500
2007-12-16 04:18:15 · answer #1 · answered by Pranil 7 · 0⤊ 0⤋
905^2-95^2 = (905+95)(905-95) = 1000*810 = 810,000
38*62+38^2 = 38(62+38) = 38(100) = 3,800
69*45 + 45*31 = 45(69+31) = 45*100 = 4500
2007-12-16 12:14:39 · answer #2 · answered by ironduke8159 7 · 0⤊ 0⤋
905^2 - 95^2
diff of two squares
(905 - 95)(905+95)
(810)(1000)=810000
(38)62+38^2
(38)(62)+(38)(38)
(38)(38+62)
38*100=3800
69*45 +45*31
45*(69+31)
45*100
4500
2007-12-16 12:21:38 · answer #3 · answered by lollyd52 2 · 0⤊ 0⤋
Not sure I would call that algebra, more like arithmetic, but I suppose you could do something like:
(905)^2 -95 ^2 -->
((1000-95)^2 -95^2 -->
(1000)^2 -2*(95)(1000) +95^2 -95^2 -->
I will leave the rest to you :)
2007-12-16 12:19:10 · answer #4 · answered by Barry C 6 · 0⤊ 0⤋