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3 years ago

What is the equation of the line that passes through (5, -2) and (-3, 4)?

2 answers:

7 0

First figure out the slope, the equation is Y2-Y1/ X2-X1.. m=-3/4 So far your equation is y=-3/4x + b. b is the y intercept. Plug in either points in the equation to have so far.. you will receive 7/4. The equation is y=-3/4 + 7/4

MatroZZZ [7]3 years ago

4 0

Answer: $4y=-3x-7$

Step-by-step explanation:

The equation of a line passing through (a,b) and (c,d) is given by :-

$(y-b)=\dfrac{d-b}{c-a}(x-a)$

The equation of a line passing through (5, -2) and (-3, 4) is given by :-

$(y-(-2))=\dfrac{4-(-2)}{-3-5}(x-5)\\\\\Rightarrow\ y+2=\dfrac{6}{-8}(x-5)\\\\\Rightarrow\ y+2=\dfrac{-3}{4}(x-5)\\\\\Rightarrow\ 4(y+2)=-3(x-5)\\\\\Rightarrow\ 4y+8=-3x+15\\\\\Rightarrow\ 4y=-3x-7$

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Consider this sphere inside the cylinder. Which statements are true? Check all that apply. NEED HELP ASAP

-Dominant- [34]

Option A: The height of the cylinder is equal to the diameter of the sphere.

Option C: The radius of the sphere is half the height of the cylinder.

Option E: The volume of the sphere is two-thirds the volume of the cylinder.

Solution:

The sphere is inside the cylinder.

Let r be the radius of the sphere.

Option A: The height of the cylinder is equal to the diameter of the sphere.

The sphere is fully occupied the cylinder.

If we draw the vertical line through enter of the sphere, which is equal to the height of cylinder.That is h = d. It is true.

Option B: The height of the cylinder is two times the diameter of the sphere.

That is h = 2d. From the above option, we know that h = d.

So, it is not true.

Option C: The radius of the sphere is half the height of the cylinder.

we know that diameter = 2 × radius (d = 2r)

From option A, we have h = d.

Substitute d = 2r, we get

⇒ h = 2r

Divide by 2 on both sides, we get

$$\Rightarrow \frac{h}{2}=r$

Therefore, it is true.

Option D: The diameter of the sphere is equal to the radius of the cylinder.

It is not true, because diameters of both cylinder and sphere are equal.

Option E: The volume of the sphere is two-thirds the volume of the cylinder.

Radius of cylinder and sphere = r

Height of cylinder h = 2r (by option C)

Volume of cylinder = Volume of sphere

$$\Rightarrow \pi r^2 h = \frac{4}{3} \pi r^3$ (formula)

$$\Rightarrow \pi r^2 (2r) = \frac{2\times2}{3} \pi r^3$

$$\Rightarrow 2 \pi r^3= \frac{2}{3} \times 2\pi r^3$

Hence it is true.

Option A, option C and option E are true.

6 0

3 years ago

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Answer:

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3 years ago

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The baby blue whale weighs 30 pounds, since 120 divided by 40 is 30.

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The Glendale Plaza Building in Glendale, California, is 353 feet tall. The U.S. Bank Tower in Los Angeles, California, is 1,017

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Answer:

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Step-by-step explanation:

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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

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(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

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

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