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Integers | Integers Exercise 1B

Question 31

20 × (-16) + 20 × 14

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Solution:-

Using the Distributive Law of Multiplication [(a×b) + (a×c) = a ×(b+c)]

Let,

a=20, b=-16, c= 14

Then,

= 20 × (-16) + 20 × 14

The above equation can be written as,

= 20 × {(-16) + 14} … [∵(a×b) + (a×c) = a ×(b+c)]

= 20 × {-2} … [∵ (+ × −= −)]

= -40

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Our top 5% students will be awarded a special scholarship to Lido.

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Question 31

20 × (-16) + 20 × 14

Looking to do well in your science exam ? Learn from an expert tutor. Book a free class!

Solution:-

Using the Distributive Law of Multiplication [(a×b) + (a×c) = a ×(b+c)]

Let,

a=20, b=-16, c= 14

Then,

= 20 × (-16) + 20 × 14

The above equation can be written as,

= 20 × {(-16) + 14} … [∵(a×b) + (a×c) = a ×(b+c)]

= 20 × {-2} … [∵ (+ × −= −)]

= -40

Our top 5% students will be awarded a special scholarship to Lido.

subject-cta
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