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3 years ago

14

How to solve arctan[tan(7pi /4)] ...?

1 answer:

Svet_ta [14]3 years ago

7 0

Well,
arctan is a bijection from R into (-pi/2 , pi/2)*and
pi is a period of tangent function:
so
as we have : tan(7pi/4) = tan(pi - pi/4) = - tan(pi/4)
we finally get :

<span>arctan(tan(7pi/4)) = artan(tan(- pi/4)) = - pi/4
</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

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A truck covered 2/7 of a journey at an average speed of 40

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Answer:

The amount of time for the whole journey is 8 hours.

Explanation:

A truck covered 2/7 of a journey at an average speed of 40 mph. Representing 1 the total of the trip traveled, then the rest of the distance traveled is calculated as: $1-\frac{2}{7} =\frac{5}{7}$

So if the truck covered the remaining 200 miles at $\frac{2}{7}$ , this means that $\frac{5}{7}$ of the trip represents the 200 miles. So, to calculate the total distance traveled by the truck, you apply the following rule of three: if $\frac{5}{7}$ of the route represents 200 miles, the integer 1 (which represents the total of the route), how many miles are they?

$miles=\frac{1*200miles}{\frac{5}{7} } =\frac{7}{5} *200 miles$

miles= 280

So the total distance traveled is 280 miles. Since speed is the relationship between the space traveled by an object and the time used for it ( $speed=\frac{distance}{time}$ ), then if the average of the entire trip was 35 mph and the distance traveled 280 miles, the time is calculated as:

$time=\frac{distance}{speed}=\frac{280 miles}{35 mph}$

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<u><em> The amount of time for the whole journey is 8 hours.</em></u>

<u><em /></u>

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3 years ago

A book, that has a mass of 0.5 grams, is pushed across a table with a force of 20 newtons. What is the acceleration of the book?

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Answer:

$4\cdot 10^4 m/s^2$

Explanation:

The acceleration of an object is given by Newton's second law:

$a=\frac{F}{m}$

where

F is the net force applied on the object

m is the mass of the object

For the book in the problem, we have:

$m=0.5 g =5\cdot 10^{-4} kg$ is the mass

$F=20 N$ is the force applied

Substituting into the formula, we find the acceleration:

$a=\frac{20 N}{5\cdot 10^{-4} kg}=4\cdot 10^4 m/s^2$

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Explanation:

It is given that,

A planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average orbit radius of Mercury.

Mass of the Sun, $M=1.99\times 10^{30}\ kg$

Radius of Mercury's orbit, $r=5.79\times 10^{10}\ m$

Radius of discovered planet, $R=\dfrac{2}{3}r$

$R=\dfrac{2}{3}\times 5.79\times 10^{10}\ m=3.86\times 10^{10}\ m$

Let T is the orbital period of such a planet. Using Kepler's third law of planetary motion as :

$T^2\propto R^3$

$T^2=\dfrac{4\pi^2R^3}{GM}$

$T^2=\dfrac{4\pi^2\times (3.86\times 10^{10})^3}{6.67\times 10^{-11}\times 1.99\times 10^{30}}$

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T = 47.86 days

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