Home
Class 8
MATHS
(- (1)/(2)) ^(3) = ?...

`(- (1)/(2)) ^(3)` = ?

A

`(-1)/(6)`

B

`1/6`

C

`1/8`

D

`(-1)/(8)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the expression \((- \frac{1}{2})^{3}\), we will follow these steps: ### Step 1: Understand the expression The expression \((- \frac{1}{2})^{3}\) means that we need to multiply \(- \frac{1}{2}\) by itself three times. ### Step 2: Write out the multiplication \[ (- \frac{1}{2})^{3} = (- \frac{1}{2}) \times (- \frac{1}{2}) \times (- \frac{1}{2}) \] ### Step 3: Multiply the first two terms First, we multiply the first two terms: \[ (- \frac{1}{2}) \times (- \frac{1}{2}) = \frac{1}{4} \] (The product of two negative numbers is positive.) ### Step 4: Multiply the result with the third term Now, we multiply \(\frac{1}{4}\) by the third term: \[ \frac{1}{4} \times (- \frac{1}{2}) = -\frac{1}{8} \] (Here, we multiply a positive number by a negative number, which results in a negative number.) ### Final Answer Thus, the value of \((- \frac{1}{2})^{3}\) is: \[ -\frac{1}{8} \] ### Summary The final answer is \(-\frac{1}{8}\). ---
Promotional Banner

Topper's Solved these Questions

  • EXPONENTS

    RS AGGARWAL|Exercise TEST PAPER-2|18 Videos
  • EXPONENTS

    RS AGGARWAL|Exercise EXERCISE 2B|7 Videos
  • DIRECT AND INVERSE PROPORTIONS

    RS AGGARWAL|Exercise TEST PAPER -12 (Fill in the blanks )|5 Videos
  • FACTORISATION

    RS AGGARWAL|Exercise EXERCISE 7E|20 Videos

Similar Questions

Explore conceptually related problems

(2)/(3) -((1)/(2) -(1)/(3))/((1)/(2)+(1)/(3))xx3(1)/(3)+(5)/(6) =?

Let H_(n)=1+(1)/(2)+(1)/(3)+ . . . . .+(1)/(n) , then the sum to n terms of the series (1^(2))/(1^(3))+(1^(2))/(1^(3))+(2^(2))/(2^(3))+(1^(2)+2^(2)+3^(2))/(1^(3)+2^(3)+3^(3))+ . . . , is

evaluate lim_ (n rarr oo) [(1) / (3) + (1) / (3 ^ (2)) + (1) / (3 ^ (2)) + ......... + (1) / (3 ^ (n))]

(((1)/(2))*((2)/(2)))/(1^(3))+(((2)/(2))*((3)/(2)))/(1^(3)+2^(3))+(((3)/(2))*((4)/(2)))/(1^(3)+2^(3)+3^(3))+...=

4 1/3-2 1/3= 2 1/3 (b) 2 3 1/3 (d) 1/2

Evaluate : 1/2 +1/3 times 1/3 div 2/3 - 1/2 times [ 1/3 + { 1/2 - ( 2/3 + 1/2 + 1/3 ) } ]

Find the sum of n-terms: [(1/1)+(1^3 +2^3)/2 +(1^3 +2^3 +3^3)/3+....to n -terms

If |{:(4,1),(2,1):}|^(2)=|{:(3,2),(1,x):}|-|{:(x,3),(-2,1):}| , then the value of x is :

Solve : (3/2)^-1 times ( 3/2 )^-1 times ( 3/2 )^-1 times ( 3/2 )^-1