What is the degree measure of an arc 4(pi) ft. long in a circle of radius 10 ft.?

2 answers:

3 0

Hey there!

$20*pi/360=4*pi/x solve for x x=360*4*pi/(20*pi)=72 degrees$

Your correct answer would be. . .

$\boxed{\boxed{72}}$

Hope this helps.
~Jurgen

Aloiza [94]3 years ago

3 0

Answer:

Hey there!

20*pi/360=4*pi/x solve for x x=360*4*pi/(20*pi)=72 degrees

Your correct answer would be. . .

Hope this helps.

Step-by-step explanation:

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Read 2 more answers

G(x) = -2x^3 – 15x^2 + 36x

shusha [124]

Consider the function $G(x) = -2x^3 - 15x^2 + 36x.$ First, factor it:

$G(x) = -2x^3 - 15x^2 + 36x=-x(2x^2+15x-36)=\\ \\=-x\cdot 2\cdot \left(x-\dfrac{-15-\sqrt{513}}{4}\right)\cdot \left(x-\dfrac{-15+\sqrt{513}}{4}\right).$

The x-intercepts are at points $\left(\dfrac{-15-\sqrt{513} }{4},0\right),\ (0,0),\ \left(\dfrac{-15+\sqrt{513} }{4},0\right).$

1. From the attached graph you can see that

function is positive for $x\in \left(-\infrty, \dfrac{-15-\sqrt{513} }{4}\right)\cup \left(0,\dfrac{-15+\sqrt{513} }{4}\right);$
function is negative for $x\in \left(\dfrac{-15-\sqrt{513} }{4},0\right)\cup \left(\dfrac{-15+\sqrt{513} }{4},\infty\right).$