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2 years ago

What is the constant rate change shown in the graph below?

Mathematics

2 answers:

Montano1993 [528]2 years ago

6 0

Answer:

the two points be

(0,0) and (1,-3)

slope [m]=[-3-0]/[1-0]=-3

constant rate is-3.

ohaa [14]2 years ago

3 0

The constant rate of change is 5 because

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A rope tied from a tent pole to a take in the ground forms 55 degrees angle with the ground. The pole is 3 feet from the stake,

grin007 [14]

Given:

A rope tied from a tent pole to a stake in the ground forms 55 degrees angle with the ground.

The pole is 3 feet from the stake.

To find:

The length of the rope to the nearest tenth of a foot.

Solution:

First draw a diagram according to the given information as shown below.

We know that, in a right angled triangle,

$\cos \theta = \dfrac{Base}{Hypotenuse}$

In triangle ABC,

$\cos 55^\circ=\dfrac{BC}{AC}$

$0.573576=\dfrac{3}{AC}$

$AC=\dfrac{3}{0.573576}$

$AC=5.23034$

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Therefore, the length of the rope is 5.2 foot.

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3 years ago

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Which expression is a fourth root of -1+isqrt3?

aleksklad [387]

Answer:

Step-by-step explanation:

$\sf n^{th}$ roots of a complex number is given by DeMoivre's formula.

$\sf \boxed{\bf r^{\frac{1}{n}}\left[Cos \dfrac{\theta + 2\pi k}{n}+i \ Sin \ \dfrac{\theta+2\pi k}{n}\right]}$

Here, k lies between 0 and (n -1) ; n is the exponent.

$\sf -1 + i\sqrt{3}$

a = -1 and b = √3

$\sf \boxed{r=\sqrt{a^2+b^2}} \ and \ \boxed{\theta = Tan^{-1} \ \dfrac{b}{a}}$

$\sf r = \sqrt{(-1)^2 + 3^2}\\\\ = \sqrt{1+9}\\\\=\sqrt{10}$

$\sf \theta = tan^{-1} \ \dfrac{\sqrt{3}}{-1}\\\\ = tan^{-1} \ (-\sqrt{3})$

$\sf = \dfrac{-\pi }{3}$

n = 4

For k = 0,

$\sf z = \sqrt[4]{10}\left[Cos \ \dfrac{\dfrac{-\pi}{3} +0}{4}+iSin \ \dfrac{\dfrac{-\pi}{3}+0}{4}\right] \\\\\\z= \sqrt[4]{10} \left[Cos \ \dfrac{ -\pi }{12}+iSin \ \dfrac{-\pi}{12}\right]\\\\\\z = \sqrt[4]{10}\left[-Cos \ \dfrac{\pi}{12}-i \ Sin \ \dfrac{\pi}{12}\right]$