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3 years ago

When give you an angle in degree and the radius, how do you calculate the arc length?

2 answers:

4 0

$\sf Hello!$

$\sf Formula\: to \:calculate \:length\: of\: an \:Arc$ :-

$\dfrac{\sf Φ}{\sf 360} × \sf πr^{2}$

Here,

• $\sf r = Radius$
• $\sf Φ = \sf Angle\: subtended\: by\: radius\: at\: the\: Centre$

~ $\sf iCarl$

Nataly [62]3 years ago

3 0

Hi there!

When radius n' angle is provided.

Use following formula to calculate th' Length of th' Arc :-

$\dfrac{\theta}{360}$ × 2πr

where,

• r = radius
• ϴ = angle subtended by radius at th' centre

~ Hope it helps

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Find all solutions to x³+3x² − 9x − 27 = 0 by factoring the equation.

SCORPION-xisa [38]

Answer:

The solutions are $x=-3,\:x=3$

Step-by-step explanation:

To factor this cubic polynomial $x^3+3x^2-9x-27$ you must:

Group the polynomial into two sections

$x^3+3x^2-9x-27=\left(x^3+3x^2\right)+\left(-9x-27\right)$

Factor out -9 from $(-9x-27)$

$(-9x-27)=-9\left(x+3\right)$

Factor out $x^2$ from $(x^3+3x^2)$

$(x^3+3x^2)=x^2\left(x+3\right)$

$x^3+3x^2-9x-27=-9\left(x+3\right)+x^2\left(x+3\right)$

Factor out common term $x+3$

$-9\left(x+3\right)+x^2\left(x+3\right)=\left(x+3\right)\left(x^2-9\right)$

$x^3+3x^2-9x-27=\left(x+3\right)\left(x^2-9\right)$

Factor $x^2-9$

$x^2-9=\left(x+3\right)\left(x-3\right)$

$x^3+3x^2-9x-27= \left(x+3\right)\left(x+3\right)\left(x-3\right)$

$x^3+3x^2-9x-27=\left(x+3\right)^2\left(x-3\right)=0$

Using the Zero Factor Theorem: = 0 if and only if = 0 or = 0

$x+3=0, \quad{x=-3}\\x-3=0, \quad{x=3}$

The solutions are

$x=-3,\:x=3$

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Step-by-step explanation:

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Answer:

D. $\frac{525\pi}{8}$ m²

Step-by-step explanation:

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Where,

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Plug in the values into the equation

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Step-by-step explanation:

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