Question

find the product of 3/5 and reciprocal of -8/15​

Answer 1

Given,

Number 1 (S1) = $3/5$

Number 2 (S2) = $- 8/15$

To Find,

Product of S1 and reciprocal of S2

Solution,

In the case of the fraction to do product, the concept is to multiply numerators with each other and denominators with each other.

So for the Fraction $S1$= $n1/d1$

and

Fraction $S2$= $n2/d2$

The product will be $(n1*n2)/(d1*d2)$

Reciprocal of the number actually means to the division of a number by 1

which means $1/s$

Finding reciprocal of $-8/15$

=$1/(-8/15)$

= $- 15/8$

Now finding the product of S1 and new S2(after reciprocal)

$= (3/5)* -(15/8)$

$= -(3*15)/(5*8)$

$= -45/40$

Hence the product of $3/5$ and reciprocal of $-8/15$ is $-45/40$