Which is an example of the commutative property?

Mathematics

2 answers:

Romashka-Z-Leto [24]3 years ago

4 0

The right answer is A

matrenka [14]3 years ago

3 0

I think B but I'm not positive.

You might be interested in

10 points. [ATTACHED]

SVETLANKA909090 [29]

Answer:

$\overline{AD}\:=\:19.2$

Step-by-step explanation:

$2x+5=3x-2\quad \:\rightarrow \quad \:x=7\\\\\overline{BD}\:=\:3\left(7\right)-2\:=\:19\\\\Length\:of\:\overline{AD}\:=\sqrt{19^2+2.5^2}=19.2$

3 0

4 years ago

Find the components of the vertical force Bold Upper FFequals=left angle 0 comma negative 4 right angle0,−4 in the directions pa

Nikolay [14]

Answer:

$F_p = < - \sqrt{3} , -3 >\\\\F_o = < \sqrt{3} , -1 >$

Step-by-step explanation:

- A plane is oriented in a Cartesian coordinate system such that it makes an angle of ( π / 3 ) with the positive x - axis.

- A force ( F ) is directed along the y-axis as a vector < 0 , - 4 >

- We are to determine the the components of force ( F ) parallel and normal to the defined plane.

- We will denote two unit vectors: ( $u_p$ ) parallel to plane and ( $u_o$ ) orthogonal to the defined plane. We will define the two unit vectors in ( x - y ) plane as follows:

- The unit vector ( $u_p$ ) parallel to the defined plane makes an angle of ( 30° ) with the positive y-axis and an angle of ( π / 3 = 60° ) with the x-axis. We will find the projection of the vector onto the x and y axes as follows:

$u_o$ = < cos ( 60° ) , cos ( 30° ) >

$u_o = < \frac{1}{2} , \frac{\sqrt{3} }{2} >$

- Similarly, the unit vector ( $u_o$ ) orthogonal to plane makes an angle of ( π / 3 ) with the positive x - axis and angle of ( π / 6 ) with the y-axis in negative direction. We will find the projection of the vector onto the x and y axes as follows:

$u_p = < cos ( \frac{\pi }{6} ) , - cos ( \frac{\pi }{3} ) >\\\\u_p = < \frac{\sqrt{3} }{2} , -\frac{1}{2} >\\$

- To find the projection of force ( F ) along and normal to the plane we will apply the dot product formulation:

- The Force vector parallel to the plane ( $F_p$ ) would be:

$F_p = u_p(F . u_p)\\\\F_p = < \frac{1}{2} , \frac{\sqrt{3} }{2} > [ < 0 , - 4 > . < \frac{1}{2} , \frac{\sqrt{3} }{2} > ]\\\\F_p = < \frac{1}{2} , \frac{\sqrt{3} }{2} > [ -2\sqrt{3} ]\\\\F_p = < -\sqrt{3} , -3 >\\$

- Similarly, to find the projection of force ( $F_o$ ) normal to the plane we again employ the dot product formulation with normal unit vector ( $u_o$ ) as follows:

$F_o = u_o ( F . u_o )\\\\F_o = < \frac{\sqrt{3} }{2} , - \frac{1}{2} > [ < 0 , - 4 > . < \frac{\sqrt{3} }{2} , - \frac{1}{2} > ] \\\\F_o = < \frac{\sqrt{3} }{2} , - \frac{1}{2} > [ 2 ] \\\\F_o = < \sqrt{3} , - 1 >$

- To prove that the projected forces ( $F_o$ ) and ( $F_p$ ) are correct we will apply the vector summation of the two orthogonal vector which must equal to the original vector < 0 , - 4 >

$F = F_o + F_p\\\\< 0 , - 4 > = < \sqrt{3}, -1 > + < -\sqrt{3}, -3 > \\\\< 0 , - 4 > = < \sqrt{3} - \sqrt{3} , -1 - 3 > \\\\< 0 , - 4 > = < 0 , - 4 >$ .. proven