${ \tt{y = {x}^{2}(x + 1) }} \\ \\ { \tt{ \frac{dy}{dx} = (2x)(x + 1) + ( {x}^{2})(1) }} \\ \\ { \tt{ \frac{dy}{dx} = 2 {x}^{2} + 2x + {x}^{2} }} \\ \\ { \tt{ \frac{dy}{dx} = 3 {x}^{2} + 2x }} \\ \\ { \boxed{ \tt{ \frac{dy}{dx} = x(3x + 2) }}}$

8 0

2 years ago

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What is the area of a regular pentagon with a side length of 10 inches and a distance from the center to a vertex of 10 inches?

Sergio [31]

Answer:

$A=172.04\ in^2$

Step-by-step explanation:

Given that,

The side length of a regular pentagon = 10 inches

The distance from the center to a vertex of 10 inches

The formula for the area of a regular pentagon is given by :

$A=\dfrac{1}{4}\sqrt{5(5+2\sqrt{5} } }\ a^2$

Put a = 10

So,

$A=\dfrac{1}{4}\times \sqrt{5(5+2\sqrt{5} )} \times 10^2\\\\A=172.04\ in^2$

So, the area of a regular pentagon is equal to $172.04\ in^2$ .

6 0

2 years ago