Question

Perform the indicated operation. Be sure the answer is reduced. d+3/8d -2d+1/10d^2

Answer 1

Answer:

$\frac{d(4d-25)}{40}$

Step-by-step explanation:

Given: $d+\frac{3}{8}d-2d+\frac{1}{10}d^2$

To find: the value of given expression in reduced form

Solution:

$d+\frac{3}{8}d-2d+\frac{1}{10}d^2=\frac{d}{1}+\frac{3}{8}d-\frac{2d}{1}+\frac{1}{10}d^2$

Here, $L.C.M(1,8,1,10)=40$

( L.C.M of two or more numbers is the smallest positive number that is their multiple )

So,

$d+\frac{3}{8}d-2d+\frac{1}{10}d^2=\frac{d}{1}+\frac{3}{8}d-\frac{2d}{1}+\frac{1}{10}d^2\\=\frac{40d+15d-80d+4d^2}{40}\\=\frac{\left ( 40d+15d-80d \right )+4d^2}{40}\\=\frac{\left ( -25d \right )+4d^2}{40}\\=\frac{4d^2-25d}{40}\\=\frac{d(4d-25)}{40}$