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प्रश्न
Find the product −3y(xy + y2) and find its value for x = 4 and y = 5.
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उत्तर
To find the product, we will use distributive law as follows:
\[- 3y\left( xy + y^2 \right)\]
\[ = - 3y \times xy + \left( - 3y \right) \times y^2 \]
\[ = - 3x y^{1 + 1} - 3 y^{1 + 2} \]
\[ = - 3x y^2 - 3 y^3\]
Substituting x = 4 and y = 5 in the result, we get:
\[- 3x y^2 - 3 y^3 \]
\[ = - 3\left( 4 \right) \left( 5 \right)^2 - 3 \left( 5 \right)^3 \]
\[ = - 3\left( 4 \right)\left( 25 \right) - 3\left( 125 \right)\]
\[ = - 300 - 375\]
\[ = - 675\]
Thus, the product is ( \[- 3x y^2 - 3 y^3\]), and its value for x = 4 and y = 5 is ( \[-\] 675).
संबंधित प्रश्न
Find each of the following product:
(−5xy) × (−3x2yz)
Find each of the following product:
\[\left( - \frac{7}{5}x y^2 z \right) \times \left( \frac{13}{3} x^2 y z^2 \right)\]
Find each of the following product:
(−4x2) × (−6xy2) × (−3yz2)
Find each of the following product:
\[\left( 0 . 5x \right) \times \left( \frac{1}{3}x y^2 z^4 \right) \times \left( 24 x^2 yz \right)\]
Express each of the following product as a monomials and verify the result in each case for x = 1:
(5x4) × (x2)3 × (2x)2
Simplify: x3y(x2 − 2x) + 2xy(x3 − x4)
Simplify: \[\frac{3}{2} x^2 ( x^2 - 1) + \frac{1}{4} x^2 ( x^2 + x) - \frac{3}{4}x( x^3 - 1)\]
Simplify:
(2x2 + 3x − 5)(3x2 − 5x + 4)
Simplify:
(3x − 2)(2x − 3) + (5x − 3)(x + 1)
What is (−4ab) × (2a²b³)?
