Math, asked by sreeja13, 1 year ago

11a3b3(7c-35)÷3a2b2(c-5)???


sreeja13: what
devashishkourav: hi
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Answers

Answered by devashishkourav
9
1 Factor out the common term 77
11a\times 3b\times 3\times \frac{7(c-5)}{3}a\times 2b\times 2(c-5)11a×3b×3×​37(c−5)​​a×2b×2(c−5)
2 Regroup terms
\frac{11\times 3\times 3\times 2\times 2}{3}aabb\times 7(c-5)(c-5)​311×3×3×2×2​​aabb×7(c−5)(c−5)
3 Simplify 11\times 3\times 3\times 2\times 211×3×3×2×2 to 396396
\frac{396}{3}aabb\times 7(c-5)(c-5)​3396​​aabb×7(c−5)(c−5)
4 Simplify \frac{396}{3}​3396​​ to 132132
132aabb\times 7(c-5)(c-5)132aabb×7(c−5)(c−5)
5 Take out the constants
(132\times 7)aabb(c-5)(c-5)(132×7)aabb(c−5)(c−5)
6 Simplify 132\times 7132×7 to 924924
924aabb(c-5)(c-5)924aabb(c−5)(c−5)
7 Use Product Rule: {x}^{a}{x}^{b}={x}^{a+b}x​a​​x​b​​=xa+b​​
924{a}^{2}{b}^{2}{(c-5)}^{2}924a​2​​b​2​​(c−5)​2​​

sreeja13: wow thanks 1
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