A puck is sitting motionless on the ice. Why does it continue to stay in one place.

SAT

1 answer:

Serjik [45]3 years ago

6 0

Answer:

Newton’s first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force. It continues to stay in one place because no force is exerted upon the puck, which means that it will continue in a line of rest.

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¿cuál de los siguientes valores de funciones es negativo? a. Sen 120° b. Sen 150° c. Cos 235° d. Cos 300°.

NeTakaya

An X-axis and Y-axis rectangular coorganizing system splits the plane in to the four regions recognised as the quadrants. The cross point is indeed the source of the two axes and is made reference to as O. So, the calculation can be defined as follows:

Given:

$\bold{a)\ \sin 120^{\circ}}\\\\\bold{b)\ \sin 150^{\circ}}\\\\\bold{c)\ \cos 235^{\circ}}\\\\\bold{d)\ \cos 300^{\circ}}\\\\$

To find:

Find negative function values =?

Solution:

$\therefore \\\\\to \bold{\sin (180- \theta)= \sin \theta} \\\\\to \bold{\cos (180- \theta)= -\cos \theta} \\\\\to \bold{\cos (180+ \theta)= -\cos \theta} \\\\ \to \bold{\cos (360- \theta)= \cos \theta}$

Solve value for a:

$\to \bold{\sin 120^{\circ}= \sin(180-60)}$

$\bold{=\sin 60^{\circ}}\\\\\bold{=\frac{\sqrt{3}}{2}}\\\\$

Solve value for b:

$\to \bold{\sin 150^{\circ}= \sin(180-30)}$

$\bold{=\sin 30^{\circ}}\\\\\bold{=\frac{1}{2}}\\\\$

Solve value for c:

$\to \bold{\cos 235^{\circ}=-\cos( 180+55)}$

$\bold{=- \cos 55^{\circ}}\\\\\bold{= -0.57357643635}}\\\\$

Solve value for d:

$\to \bold{\cos 300^{\circ}= \cos(360-60)}$

$\bold{= \cos 60^{\circ}}\\\\\bold{= \frac{1}{2}}\\\\$

Therefore the final answer is "Option C".

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brainly.com/question/24581443

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3 years ago

17.1 equation of a circle evaluate homework and practice answer key

sveta [45]

The evaluation using the equation of circle for the practice question is (x - 4)² + (y + 11)² = 256

<h3>What is the equation of the Circle Distance Formula?</h3>

With respect to the distance formula, a point location in space (x,y) at a distance (r) from the origin is valid if $\mathbf{\sqrt{x^2 + y^2} = r}$ . Hence this refers to the equation of the circle centered at the origin.