I will give BRAINLIEST and 20 points answer it.

Mathematics

1 answer:

Lostsunrise [7]3 years ago

6 0

Answer:

Step-by-step explanation:

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Could you please help me!?

Alinara [238K]

Answer:

Answer: 42.5

Step-by-step explanation:

It says 1 inch equals 5 feet. So, every inch is on the map is an extra 5 feet in real space. So if you think about it, all this is is a simple multiplication problem.

Here is the math:

8.5 * 5 = 42.5

Think about it, for every inch there is on the map there is an extra 5 feet in real space. There is 8 inches, so we need 5 extra feet for each of those 8 inches. So what we get for 1 inch is 5 feet, for 2 inches 10 feet, 3 inches 15 feet, etc etc until we reach 8 inches. For the 0.5 inches alone that would equal 2.5 feet in real life since 1 inch is 5 feet.

Ez pz

6 0

3 years ago

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

Gennadij [26K]

(a)

$\displaystyle \frac{1}{2} \cdot \int_{\frac{\pi}{3}}^{\frac{5\pi}{3}} \left(4^2 - (3 + 2\cos\theta)^2 \right) \, d\theta$

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
r = 3 + 2cosθ, it follows that

$x = (3 + 2\cos\theta) \cos \theta \\ 
y = (3 + 2\cos\theta) \sin \theta$

Differentiating with respect to theta

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

2/3 is the slope

____________

(c)

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

Angle θ and r are related via r = 3 + 2cosθ, so implicitly differentiating with respect to time

 $\displaystyle\frac{dr}{dt} = -2\sin\theta \frac{d\theta}{dt} \quad \stackrel{\theta = \pi/3}{\implies} \quad 3 = -2\left( \frac{\sqrt{3}}{2}}\right) \frac{d\theta}{dt} \implies \\ \\ \frac{d\theta}{dt} = -\sqrt{3} \text{ radians per second}$