Answer:
[tex]\frac{d(4d-25)}{40}[/tex]
Step-by-step explanation:
Given: [tex]d+\frac{3}{8}d-2d+\frac{1}{10}d^2[/tex]
To find: the value of given expression in reduced form
Solution:
[tex]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[/tex]
Here, [tex]L.C.M(1,8,1,10)=40[/tex]
( L.C.M of two or more numbers is the smallest positive number that is their multiple )
So,
[tex]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}[/tex]
