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2 years ago

Find the coefficient of the 4th term in expansion of (x +2 )^5

1 answer:

6 0

$(x + 2)^5=(x+2)^2\cdot{(x+2)^2}\cdot{(x+2)}\\\\(x+2)^5=(x^2+4x+4)\cdot{(x^2+4x+4)}\cdot(x+2)$

$(x+2)^5=(x^4+8x^3+24x^2+32x+16)\cdot(x+2)$

$(x+2)^5=x^5+10x^4+40x^3+80x^2+80x+32$

10

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3 years ago

A triangle has a base of (3x + 7) and a height of (5x - 1). A second

Mars2501 [29]

Answer:

The difference between the area of the original triangle and the area of the new triangle is $\Delta A_{\bigtriangleup} = \frac{5}{2}\cdot (3\cdot x +7)\cdot (5\cdot x -1)$ .

Step-by-step explanation:

The equation for the area of a triangle ( $A_{\bigtriangleup}$ ) is:

$A_{\bigtriangleup} = \frac{1}{2}\cdot b \cdot h$

Where:

$b$ - Base, dimensionless.

$h$ - Height, dimensionless.

The expression for each triangle are described below:

First Triangle ( $b = 3\cdot x + 7$ , $h = 5\cdot x - 1$ )

$A_{\bigtriangleup,1} = \frac{1}{2}\cdot (3\cdot x+7)\cdot (5\cdot x -1)$

Second Triangle ( $b = 3\cdot (3\cdot x+7)$ , $h = 2\cdot (5\cdot x -1)$ )

$A_{\bigtriangleup,2} = 3\cdot (3\cdot x+7)\cdot (5\cdot x -1)$

The difference between the area of the original triangle and the area of the new triangle is:

$\Delta A_{\bigtriangleup} = A_{\bigtriangleup,2}-A_{\bigtriangleup,1}$

$\Delta A_{\bigtriangleup} = 3\cdot (3\cdot x+7)\cdot (5\cdot x-1)-\frac{1}{2} \cdot (3\cdot x+7)\cdot (5\cdot x-1)$

$\Delta A_{\bigtriangleup} = \frac{5}{2}\cdot (3\cdot x +7)\cdot (5\cdot x -1)$

The difference between the area of the original triangle and the area of the new triangle is $\Delta A_{\bigtriangleup} = \frac{5}{2}\cdot (3\cdot x +7)\cdot (5\cdot x -1)$ .

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