Question

(2/3×[(4/2)×(-21/4)]​

Answer 1

Answer:

 $(\frac{2}{3} [(\frac{4}{2}) *(\frac{-21}{4})] = -\frac{29}{6}$    

Step-by-step explanation:

Given : The given expression is $(\frac{2}{3} [(\frac{4}{2}) *(\frac{-21}{4})]$

To find : The value of this given expression.

Solution :

• The given expression is $(\frac{2}{3} [(\frac{4}{2}) *(\frac{-21}{4})]$
• We have to solve this expression and find out final value.
• To solve this expression, we have to use BODMAS Rule. It states that, while solving an any expression, we should first solve B-Brackets, O-Order, D-Division, M-Multiplication, A-Addition and S-Subtraction.
• By following BODMAS Rule, the expression becomes,

      $(\frac{2}{3} [(\frac{4}{2}) *(\frac{-21}{4})]$ = $\frac{2}{3} *[(\frac{-21*4}{8} )]$

                         = $\frac{2}{3}*(-\frac{84}{8})$

                         = $-\frac{2*84}{3*8}$

                         = $\frac{-116}{24}$

                         = $-\frac{29}{6}$

• ∴  $(\frac{2}{3} [(\frac{4}{2}) *(\frac{-21}{4})] = -\frac{29}{6}$    

Answer 2

Given: The equation is$\frac{2}{3} \times[(\frac{4}{2} )\times(\frac{-21}{4} )]$

We have to find the value of the above equation.

By using the Bodmas rule, we are solving the above equation.

As we know that the Bodmas rule is the rule used to remember the order of operations to be followed while solving expressions in mathematics.

where,

$\begin{array}{l}\mathrm{B}=\text{brackets}\\\mathrm{O}=\text { order of powers or rules } \\\mathrm{D}=\text { division } \\\mathrm{M}=\text { multiplication } \\\mathrm{A}=\text { addition } \\\mathrm{S}=\text { subtraction }\end{array}$

We are solving in the following way:

We have,

The equation is$\frac{2}{3} \times[(\frac{4}{2} )\times(\frac{-21}{4} )]$

$=>\frac{2}{3} \times[2\times(-5.25)]\\\\=>\frac{2}{3} \times[-10.5]\\\\=>-7$

Hence, after solving the above equation we get, the solution of the equation is $-7.$