Question

11. 14(3y - 52)+7(3y - 52)
Factories the following polynomial​

Answer 1

Answer࿐

Find two consecutive odd numbers such that two fifths of the smaller number exceeds two ninths of the larger by 4

Solution :

Let one odd number be ' 2n + 1 '

This is smallest odd number .

Other consecutive odd number be ' 2n + 3 '

This is largest odd number .

A/c , " Two fifths of the smaller number exceeds two ninths of the larger by 4 "

First consecutive smallest odd number :

= 2n + 1

= 2(12) + 1

= 24 + 1

= 25

Second consecutive largest odd number :

= 2n + 3

= 2(12) + 3

= 24 + 3

= 27

Alternative : You may solve this question by taking ' x ' as smallest consecutive odd number and ' x + 2 ' as biggest consecutive odd number .

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Answer 2

Answer:

(3y-52)(14+7)

21(3y-52)

Step-by-step explanation:

(3y-52) is common term we will take it in the first place

(14+7) is the uncommon term which when added become 21 so, 21 will come first then the expression

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