Question

Pls Tell how To Do This Question​

Answer 1

-11/28

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Answer 2

Answer:

Given:-

Solve : $\dfrac{2}{5} × ( \dfrac{-3}{7} ) - \dfrac{1}{6} × \dfrac{3}{2} + \dfrac{1}{14} × \dfrac{2}{5}$ .

To Find:-

Solve the given question.

Note:-

●》Here, we will solve the sum by BODMAS method. According to BODMAS, we will first solve "bracket value"; Next "of" value: next "divisional value ÷ "; Next "multiple value ×"; Next "addition value +"; At last, "subtraction value -".

Solution:-

$\huge\red{ \dfrac{2}{5} × ( \dfrac{-3}{7} ) - \dfrac{1}{6} × \dfrac{3}{2} + \dfrac{1}{14} × \dfrac{2}{5}}$

$\huge\red{ \ \ Their \ \ solution = ?}$

☆ According to note first point~

▪︎$\dfrac{2}{5} × ( \dfrac{-3}{7} ) - \dfrac{1}{6} × \dfrac{3}{2} + \dfrac{1}{14} × \dfrac{2}{5}$

☆ In brackets, there is no value to solve, so we will simply open it~

▪︎$\dfrac{2}{5} × \dfrac{-3}{7} - \cancel\dfrac{1}{6} × \cancel\dfrac{3}{2} + \cancel\dfrac{1}{14} × \cancel\dfrac{2}{5}$

▪︎$\dfrac{-6}{35} - \dfrac{1}{4} + \dfrac{1}{35}$

☆ L.C.M of denominators 35, 4, 35 = 140~

▪︎$\dfrac{-6}{35} × \dfrac{4}{4} - \dfrac{1}{4} × \dfrac{35}{35} + \dfrac{1}{35} × \dfrac{4}{4}$

▪︎$\dfrac{-24}{140} - \dfrac{35}{140} + \dfrac{4}{140}$

▪︎$\dfrac{-24}{140} - \dfrac{31}{140}$

▪︎$\dfrac{-55}{140}$

☆ After reducing by 5~

▪︎$\dfrac{-11}{28}$

$\huge\pink{Their \ \ solution = \dfrac{-11}{28}}$

Answer:-

Hence, the solution of $\dfrac{2}{5} × ( \dfrac{-3}{7} ) - \dfrac{1}{6} × \dfrac{3}{2} + \dfrac{1}{14} × \dfrac{2}{5} \implies \dfrac{-11}{28}$ .

:)