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3 years ago

What is the eqaution of a vertical line passing through (-5,-2)?

Mathematics

1 answer:

Virty [35]3 years ago

8 0

Answer:

x=-5

Step-by-step explanation:

Vertical lines have the same x value all the time. They are of the form x=

Since it passes through the point (-5,-2) the x value must be -5

x=-5

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Answer:

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Step-by-step explanation:

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3 years ago

Consider the infinite geometric series ∑∞n=1−4(13)n−1

vodomira [7]

Answer:

See below

Step-by-step explanation:

-4 (1/3)^(n-1)

Part A

For n = 1

-4(1/3)^(1 - 1)

-4(1/3)^0 Anything to the 0 power (except 0) is 1

-4 (1)

-4

n = 2

-4(1/3)^(2 - 1)

-4*(1/3)^1

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n = 3

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Part B

The series converges.

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Part C

Formula

Sum = a/(1 - r)

a = - 4

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3 years ago

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AnnZ [28]

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2 years ago

Mg+rg=2H, solve for g

lesantik [10]

Answer:

g = 2H/(m + r)

Step-by-step explanation:

mg + rg = 2H

g(m + r) = 2H

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g = 2H/(m + r)

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3 years ago

You are given the parametric equations x=2cos(θ),y=sin(2θ). (a) List all of the points (x,y) where the tangent line is horizonta

vladimir1956 [14]

Answer:

The solutions listed from the smallest to the greatest are:

x: $-\sqrt{2}$ $-\sqrt{2}$ $\sqrt{2}$ $\sqrt{2}$

y: -1 1 -1 1

Step-by-step explanation:

The slope of the tangent line at a point of the curve is:

$m = \frac{\frac{dy}{dt} }{\frac{dx}{dt} }$

$m = -\frac{\cos 2\theta}{\sin \theta}$

The tangent line is horizontal when $m = 0$ . Then:

$\cos 2\theta = 0$

$2\theta = \cos^{-1}0$

$\theta = \frac{1}{2}\cdot \cos^{-1} 0$

$\theta = \frac{1}{2}\cdot \left(\frac{\pi}{2}+i\cdot \pi \right)$ , for all $i \in \mathbb{N}_{O}$

$\theta = \frac{\pi}{4} + i\cdot \frac{\pi}{2}$ , for all $i \in \mathbb{N}_{O}$

The first four solutions are:

x: $\sqrt{2}$ $-\sqrt{2}$ $-\sqrt{2}$ $\sqrt{2}$

y: 1 -1 1 -1

The solutions listed from the smallest to the greatest are:

x: $-\sqrt{2}$ $-\sqrt{2}$ $\sqrt{2}$ $\sqrt{2}$

y: -1 1 -1 1

6 0

2 years ago

What is the eqaution of a vertical line passing through (-5,-2)?​

What is the eqaution of a vertical line passing through (-5,-2)?