What is expand 4x(y^1f(x,y,O)+2xy^1-y=O) ?

The solution to expand 4x(y^1f(x,y,O)+2xy^1-y=O) is 4xy^1f(,,xyO)+4*2xxy^1-4xy=O

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

expand 4x(y^1f(x,y,O)+2xy^1-y=O)

expand

4 x (y^{1} f (x, y, O) + 2 x y^{1} - y = O)

4 x y^{1} f (,, x y O) + 4 \cdot 2 xx y^{1} - 4 x y = O

Solution steps

4 x (y^{1} f (x, y, O) + 2 x y^{1} - y = O)

= 4 x (y^{1} f (,, x y O) + 2 x y^{1} - = y O)

Distribute parentheses

= 4 x y^{1} f (x, y, O) + 4 x \cdot 2 x y^{1} + 4 x (- y = O)

Apply minus-plus rules

+ (- a) = - a

4 x y^{1} f (,, x y O) + 4 \cdot 2 xx y^{1} - 4 x y = O

\frac{g ( x + h ) - 4}{g ( x + h ) + 5}

\frac{2}{x ( x + 3 )}

2.3 (x + 1.4 = - 9.66)

\frac{( a + b ) ^{12}}{( a - b ) ^{10}}

\frac{9 ( x + 9 )}{x}

What is expand 4x(y^1f(x,y,O)+2xy^1-y=O) ?
The solution to expand 4x(y^1f(x,y,O)+2xy^1-y=O) is 4xy^1f(,,xyO)+4*2xxy^1-4xy=O