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3 years ago

I really need help!!!!!!!!

Mathematics

2 answers:

jeka943 years ago

6 0

Answer: -1.25 < -2/3 < 2.75 < 4 < 9/2

Step-by-step explanation: hope it helps

Yakvenalex [24]3 years ago

5 0

Answer:

least to greatest- -1.25, -2/3, 2.75, 4, 9/2

Step-by-step explanation:

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A line passes through the points (1, –6) and (4, 3).

Marina CMI [18]

Answer:

y = 3x - 9

Step-by-step explanation:

We first need to find the slope (m)

m = $\frac{y_2-y_1}{x_2-x_1}$ = $\frac{3 - (-6)}{4-1}$ = $\frac{9}{3}$ = 3

y = mx + b

y = 3x + b

(4, 3)

y = 3x + b

3 = 3(4) + b

3 = 12 + b

-12 -12

-9 = b

(1, –6)

y = 3x + b

-6 = 3(1) + b

-6 = 3 + b

-3 -3

-9 = b

y = 3x - 9

To check: (1, –6)

y = 3x - 9

-6 = 3(1) - 9

-6 = 3 - 9

-6 = -6

Hope this helps!

6 0

2 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

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