8(3/16)+2(6 1/4)-13/16<br> How to solve

How do you solve -26d = -364?

Novay_Z [31]

You simply divide -364 by -26 so d=14
hope this helps! :D

6 0

2 years ago

Read 2 more answers

How do you find velocity of a dt graph

givi [52]

Answer: to find velocity, simply put (d2-d1) over (t2-t1) or d/t.

Explanation: To find velocity, divide the change in distance by the change in time. On a graph, pick two points and find the slope of one portion of the line.
Why the slope?
Well this is like your normal x vs. y graph. X being the time and y being your distance.
To find the slope for this, it uses the same formula: (Y2-y1) / (x2-x1)

Hope this helps!!!
BTW, this is physics.

6 0

2 years ago

50 POINTS AND BRAINLIEST!!!! PLEASE HELP IT'S IMPORTANT!!!!!

zvonat [6]

Answer:

p=7

q=9

r=13

Step-by-step explanation:

7 0

2 years ago

What are the magnitude and direction of u+v+w ?

nevsk [136]

The magnitude and the direction of the resultant are approximately 57.871 and 198.676°.

<h3>How to determine the magnitude and the direction of a resultant</h3>

Vectors are elements with two given characteristics: Magnitude and direction. A resultant is derived from the sum of vectors, the magnitude is the norm of a vector and the direction is the orientation of a vector. The resultant and its characteristics are described below:

<h3>Resultant</h3>

$\vec r = \left(\sum\limits_{i=1}^{n}x_{i}, \sum\limits_{i=1}^{n}y_{i}\right)$ (1)

<h3>Magnitude</h3>

$\|\vec r\| = \sqrt{\left(\sum\limits_{i=1}^{n} x_{i}\right)^{2}+\left(\sum\limits_{i=1}^{n} y_{i}\right)^{2}}$ (2)

<h3>Direction</h3>

$\theta = \tan^{-1}\frac{\sum\limits_{i=1}^{n}y_{i}}{\sum\limits_{i=1}^{n}x_{i}}$ (3)

Where $\theta$ is resultant angle, measured in degrees.

If we know that $\|\vec u\| = 90$ , $\theta_{u} = 40^{\circ}$ , $\|\vec v\| = 60$ , $\theta_{v} = 20^{\circ}$ , $\|\vec w\| = 40$ and $\theta_{w} = 30^{\circ}$ , then the resultant, its magnitude and its direction:

<h3>Resultant</h3>

$\vec r = (-90\cdot \cos 40^{\circ}-60\cdot \sin 20^{\circ}+40\cdot \cos 30^{\circ}, 40\cdot \sin 30^{\circ}+60\cdot \cos 20^{\circ}-90\cdot \sin 40^{\circ})$