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mariarad [96]
3 years ago
14

What is the sum of the positive odd divisors of 60?

Mathematics
2 answers:
GREYUIT [131]3 years ago
7 0

Answer:168 also the factors are 1,2,3,4,5,6,10,12,15,20,30

Step-by-step explanation:simple logic answer☀️

posledela3 years ago
7 0
She or he is right that’s the correct answer
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A hula hoop has an circumference 150.72 cm. What is the area of the hula hoop? (round to the nearest hundredth)
Leya [2.2K]

Answer:

Area = 1808.64 cm

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3 years ago
Given the equation of the circle, identify the coordinates of the center. (x+2)2+(y-6)2=49
sveta [45]

Answer:

(-2, 6)

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Equation of a circle is:

(x − h)² + (y − k)² = r²

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The graphs of the polar curves r = 4 and r = 3 + 2cosθ are shown in the figure above. The curves intersect at θ = π/3 and θ = 5π
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\displaystyle \frac{1}{2} \cdot \int_{\frac{\pi}{3}}^{\frac{5\pi}{3}} \left(4^2 - (3 + 2\cos\theta)^2 \right) \, d\theta

or, via symmetry

\displaystyle\frac{1}{2} \cdot 2 \int_{\frac{\pi}{3}}^{\pi} \left(4^2 - (3 + 2\cos\theta)^2 \right) \, d\theta

____________

(b)

By the chain rule:

\displaystyle \frac{dy}{dx} = \frac{ dy/ d\theta}{ dx/ d\theta}

For polar coordinates, x = rcosθ and y = rsinθ. Since
<span>r = 3 + 2cosθ, it follows that

x = (3 + 2\cos\theta) \cos \theta \\ &#10;y = (3 + 2\cos\theta) \sin \theta

Differentiating with respect to theta

\begin{aligned}&#10;\displaystyle \frac{dy}{dx} &= \frac{ dy/ d\theta}{ dx/ d\theta} \\&#10;&= \frac{(3 + 2\cos\theta)(\cos\theta) + (-2\sin\theta)(\sin\theta)}{(3 + 2\cos\theta)(-\sin\theta) + (-2\sin\theta)(\cos\theta)} \\ \\&#10;\left.\frac{dy}{dx}\right_{\theta = \frac{\pi}{2}}&#10;&= 2/3&#10;\end{aligned}

2/3 is the slope

____________

(c)

"</span><span>distance between the particle and the origin increases at a constant rate of 3 units per second" implies dr/dt = 3

A</span>ngle θ and r are related via <span>r = 3 + 2cosθ, so implicitly differentiating with respect to time

</span><span />\displaystyle\frac{dr}{dt} = -2\sin\theta \frac{d\theta}{dt} \quad \stackrel{\theta = \pi/3}{\implies} \quad 3 = -2\left( \frac{\sqrt{3}}{2}}\right) \frac{d\theta}{dt} \implies \\ \\ \frac{d\theta}{dt} = -\sqrt{3} \text{ radians per second}
5 0
3 years ago
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