Mathematics
Grade6
Easy

Question

Evaluate ( - 4a2 ), where a = - 4.

  1. -64    
  2. 64    
  3. 16    
  4. -16    

Hint:

This is an algebraic expression. An algebraic expression consists of two components: a variable and a constant. A variable is a quantity that doesn’t have any fixed value. It takes value according to the conditions. It is denoted by the alphabet. A constant is a quantity with a fixed value. In this question, value is assigned to the variables. We have to perform the required operations and find the final value.

The correct answer is: -64


    The given expression is (-4a2)
    There is one variable and one constant in the above expression. The variable is: ‘a’. Value assigned to the variable is a = -4. We have to substitute this value and find the final value of the expression.
    There is one term in the above expression. The term … means we have to multiply the variable with itself. And the multiply the whole value by (-4).
    We have to first substitute the value and perform the required operations.
    To substitute the value, we have to replace a by -4. After substituting the value, we will get the following equation:
    (-4a2) = [-4(-4)2]
    First we will square the (-4).
    (-4a2) = (-4 × 16)
    = -64
    Therefore, the value of equation is -64.
    So option (d), which is ‘-64’, is the right option.

    We have to be careful about the operations because of the variables. We have to follow the proper procedure for performing the operations. The variables take different values under different conditions. So we have to be careful while substituting the values. The main part of this type of question is just replacing the variables by their values and performing the operation.

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