Sum of three consecutive multiples of 6 is 648. One of these multiple is​

Let two consecutive multiples of 3 be x and (x+3)

A/q,

x * (x+3) = 648

➡ x² + 3x = 648

➡ x² + 3x -648 = 0

➡ x² + 27x - 24x -648 = 0

➡ x ( x + 27 ) -24 ( x +27)

➡ ( x - 24) ( x + 27)

➡ x = 24 and x = -27

so, we take x = 24.

Required multiples of 3

➡ x = 24

➡ x +3 = 24+3 = 27.

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b) 1/3 * 1/5

step-by-step explanation:

each smallest part is 1/15 since the written one is 1/5 and it's divided by 3 equal parts. (1/5 )/3 = 1/15 = 1/3 * 1/5