Home
Class 8
MATHS
9 x^(6) y^(2) ÷ 3x^(3) y =...

`9 x^(6) y^(2) ÷ 3x^(3) y =`

A

`3x^(2) y^(2)`

B

`3x^(3) y`

C

`3x^(3) y^(2)`

D

`6x^(3) y`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the expression \( 9x^6y^2 \div 3x^3y \), we will follow these steps: ### Step 1: Write the expression clearly We start with the expression: \[ \frac{9x^6y^2}{3x^3y} \] ### Step 2: Simplify the coefficients First, we simplify the numerical coefficients (the numbers in front): \[ \frac{9}{3} = 3 \] So, we can rewrite the expression as: \[ 3 \cdot \frac{x^6y^2}{x^3y} \] ### Step 3: Simplify the variable \(x\) Next, we simplify the \(x\) terms. According to the exponent rule for division, which states that \(a^m \div a^n = a^{m-n}\): \[ \frac{x^6}{x^3} = x^{6-3} = x^3 \] So, we have: \[ 3 \cdot x^3 \cdot \frac{y^2}{y} \] ### Step 4: Simplify the variable \(y\) Now, we simplify the \(y\) terms using the same exponent rule: \[ \frac{y^2}{y} = y^{2-1} = y^1 = y \] Thus, the expression now becomes: \[ 3x^3y \] ### Final Answer The final simplified expression is: \[ 3x^3y \] ---
Promotional Banner

Topper's Solved these Questions

  • EXPONENTS AND POWERS

    MTG IIT JEE FOUNDATION|Exercise Exercise (Multiple Choice Questions) LEVEL - 2|15 Videos
  • EXPONENTS AND POWERS

    MTG IIT JEE FOUNDATION|Exercise Exercise (Match the Following)|3 Videos
  • EXPONENTS AND POWERS

    MTG IIT JEE FOUNDATION|Exercise NCERT Section (Exercise 12.2)|17 Videos
  • DIRECT AND INVERSE PROPORTIONS

    MTG IIT JEE FOUNDATION|Exercise OLYMPIAD/HOTS CORNER|15 Videos
  • FACTORISATION

    MTG IIT JEE FOUNDATION|Exercise OLYMPIAD/HOTS CORNER|10 Videos

Similar Questions

Explore conceptually related problems

Find the centre and radius of the circle 3x^(2) 3y^(2) - 6x +9y - 8 =0 .

The HCF of 4y^(4)x - 9y^(2)x^(3) and 4y^(2)x^(2) + 6yx^(3) is

Find each of the following products: (i) (x - 4)(x - 4) (ii) (2x - 3y)(2x - 3y) (iii) ((3)/(4) x - (5)/(6) y) ((3)/(4)x - (5)/(6) y) (iv) (x - (3)/(x)) (x - (3)/(x)) (v) ((1)/(3) x^(2) - 9) ((1)/(3) x^(2) - 9) (vi) ((1)/(2) y^(2) - (1)/(3) y) ((1)/(2) y^(2) - (1)/(3) y)

The HCF of 4y^(4)x-9y^(2)x^(3) and 4y^(2)x^(2)+6yx^(3) is

If the equation 3x^(2) + 3y^(2) + kxy + 9x + (k - 6) y + 3 = 0 represents a circle , then the radius of this circle is

Let |{:(y^(5)z^(6)(z^(3)-y^(3)),,x^(4)z^(6)(x^(3)-z^(3)),,x^(4)y^(5)(y^(3)-x^(3))),(y^(2)z^(3)(y^(6)-z^(6)),,xz^(3)(z^(6)-x^(6)) ,,xy^(2)(x^(6)-y^(6))),(y^(2)^(3)(z^(3)-y^(3)),,xz^(3)(x^(3)-z^(3)),,xy^(2)(y^(3)-x^(3))):}| " and " Delta_(2)= |{:(x,,y^(2),,z^(3)),(x^(4),,y^(5) ,,z^(6)),(x^(7),,y^(8),,z^(9)):}| .Then Delta_(1)Delta_(2) is equal to

{:("Column" A ,, "Column" B), ((3x^(2) - 5)- (2x^(2) - 5 + y^(2)) ,, (a) x^(2) + xy + y^(2)) , (9x^(2) - 16y^(2) ,, (b) 2) , ((x^(3) - y^(3))/(x-y) ,, (c) (9x + 16y) (9x - 16y)) , ("The degree of " (x + 2) (x+3) ,, (d) x^(2) - y^(2)) , (,, (e) 1) , (,, (f) (3x + 4y) (3x - 4y)):}

If 9x^(2)+25y^(2)=181 and xy=-6, find the value of 3x+5y