Answer "yes" or "no" and show you work

Mathematics

1 answer:

goblinko [34]3 years ago

7 0

Answer:

Yes

Step-by-step explanation:

y = -8x + 5

x = -8y + 5

x - 5 = -8y

x/-8 -5/-8 = y

y = -x/8 + 5/8

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Graph a line that contains the point (4, 3) and has a slope of 1/2

Lapatulllka [165]

Answer:

y=1/2x+1

Step-by-step explanation:

one dot on the 1 on the y axis

then more one unit up and two units to the left

and then from the first point (0,1) go one unit down

and two to the right. then make the line with the three points

3 0

3 years ago

Assume the mean useful life of a particular light bulb is 2,000 hours and is normally distributed with a standard deviation of 3

Yuri [45]

The mean = 2,000
s = Sigma ( standard deviation ) = 300
M - 2 s = 2,000- 2 * 300 = 2,000 - 600 = 1,400
M + 2 s = 2,000 + 2 * 300 = 2,000 + 600 = 2,600
Answer.
Useful life of light bulbs within 2 standard deviations of the mean is:
A ) Between 1,400 and 2,600 hours

4 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

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