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3 years ago

Does the table represent a proportional relationship

Mathematics

1 answer:

AleksandrR [38]3 years ago

7 0

Answer:no

Step-by-step explanation: No, because all the ratios of y to x are not equal.

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How to find the greatest common factor of 24 and 90?

love history [14]

Well, they're both divisible by 2 (24 ÷ 2 = 12) (90 ÷ 2 = 45)

3 0

3 years ago

Given the domain (-2,2,4) what is the range for the relation 3x-y=3

Bumek [7]

Answer:

Given the domain, the range for 3x-y = 3 is {-9, 6, 9}

Step-by-step explanation:

First you have to put the relation in terms of y ⇒ 3x - y = 3⇒ 3x -3 = y

⇒ y = 3x - 3.

Then you replace the values indicated by the domain to find their "y" values (the ones that constitute the range).

f(-2) = -9

f(2) = 6

f(4) = 9.

Finally, the range for the given domain is {-9, 6, 9}

3 0

3 years ago

What is the scientific notation of 6174798604

lesantik [10]

Here is your answer:

Since their no zeros (0) here's your answer:

6.174798604×10^9

7 0

3 years ago

The circumference of the

Tju [1.3M]

To the nearest inch it'll be

d×3.14=63

= 20 inches

8 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

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

5 0

3 years ago