Question

1. 4a (3α +7b) . find each of the following products:
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Answer 1

Answer:

By distributive law of multiplication over addition, we have:

By distributive law of multiplication over addition, we have:p×(q+r)=(p×q)+(p×r)

By distributive law of multiplication over addition, we have:p×(q+r)=(p×q)+(p×r)where p,q and r be three monomials.

By distributive law of multiplication over addition, we have:p×(q+r)=(p×q)+(p×r)where p,q and r be three monomials.∴ 4a(3a+7b)

By distributive law of multiplication over addition, we have:p×(q+r)=(p×q)+(p×r)where p,q and r be three monomials.∴ 4a(3a+7b)=(4a×3a)+(4a×7b)

By distributive law of multiplication over addition, we have:p×(q+r)=(p×q)+(p×r)where p,q and r be three monomials.∴ 4a(3a+7b)=(4a×3a)+(4a×7b)=(12a

By distributive law of multiplication over addition, we have:p×(q+r)=(p×q)+(p×r)where p,q and r be three monomials.∴ 4a(3a+7b)=(4a×3a)+(4a×7b)=(12a 1+1

By distributive law of multiplication over addition, we have:p×(q+r)=(p×q)+(p×r)where p,q and r be three monomials.∴ 4a(3a+7b)=(4a×3a)+(4a×7b)=(12a 1+1 +28ab) [∵a

By distributive law of multiplication over addition, we have:p×(q+r)=(p×q)+(p×r)where p,q and r be three monomials.∴ 4a(3a+7b)=(4a×3a)+(4a×7b)=(12a 1+1 +28ab) [∵a m

By distributive law of multiplication over addition, we have:p×(q+r)=(p×q)+(p×r)where p,q and r be three monomials.∴ 4a(3a+7b)=(4a×3a)+(4a×7b)=(12a 1+1 +28ab) [∵a m ×a

By distributive law of multiplication over addition, we have:p×(q+r)=(p×q)+(p×r)where p,q and r be three monomials.∴ 4a(3a+7b)=(4a×3a)+(4a×7b)=(12a 1+1 +28ab) [∵a m ×a n

By distributive law of multiplication over addition, we have:p×(q+r)=(p×q)+(p×r)where p,q and r be three monomials.∴ 4a(3a+7b)=(4a×3a)+(4a×7b)=(12a 1+1 +28ab) [∵a m ×a n =a

By distributive law of multiplication over addition, we have:p×(q+r)=(p×q)+(p×r)where p,q and r be three monomials.∴ 4a(3a+7b)=(4a×3a)+(4a×7b)=(12a 1+1 +28ab) [∵a m ×a n =a m+n

By distributive law of multiplication over addition, we have:p×(q+r)=(p×q)+(p×r)where p,q and r be three monomials.∴ 4a(3a+7b)=(4a×3a)+(4a×7b)=(12a 1+1 +28ab) [∵a m ×a n =a m+n ]

By distributive law of multiplication over addition, we have:p×(q+r)=(p×q)+(p×r)where p,q and r be three monomials.∴ 4a(3a+7b)=(4a×3a)+(4a×7b)=(12a 1+1 +28ab) [∵a m ×a n =a m+n ]=(12a

By distributive law of multiplication over addition, we have:p×(q+r)=(p×q)+(p×r)where p,q and r be three monomials.∴ 4a(3a+7b)=(4a×3a)+(4a×7b)=(12a 1+1 +28ab) [∵a m ×a n =a m+n ]=(12a 2

By distributive law of multiplication over addition, we have:p×(q+r)=(p×q)+(p×r)where p,q and r be three monomials.∴ 4a(3a+7b)=(4a×3a)+(4a×7b)=(12a 1+1 +28ab) [∵a m ×a n =a m+n ]=(12a 2 +28ab)

By distributive law of multiplication over addition, we have:p×(q+r)=(p×q)+(p×r)where p,q and r be three monomials.∴ 4a(3a+7b)=(4a×3a)+(4a×7b)=(12a 1+1 +28ab) [∵a m ×a n =a m+n ]=(12a 2 +28ab)Hence, 4a(3a+7b)=12a

By distributive law of multiplication over addition, we have:p×(q+r)=(p×q)+(p×r)where p,q and r be three monomials.∴ 4a(3a+7b)=(4a×3a)+(4a×7b)=(12a 1+1 +28ab) [∵a m ×a n =a m+n ]=(12a 2 +28ab)Hence, 4a(3a+7b)=12a 2

By distributive law of multiplication over addition, we have:p×(q+r)=(p×q)+(p×r)where p,q and r be three monomials.∴ 4a(3a+7b)=(4a×3a)+(4a×7b)=(12a 1+1 +28ab) [∵a m ×a n =a m+n ]=(12a 2 +28ab)Hence, 4a(3a+7b)=12a 2 +28ab.

Answer 2

Step-by-step explanation:

$4a(3a + 7b) \\ = 12 {a}^{2} + 28ab$