One way to determine if a square root is rational or irrational is

Mathematics

1 answer:

sasho [114]3 years ago

7 0

Answer:

I thought that all races were = I thought it was fine we all loved each other and that there were protests. That the black lives matter. if a police officer kills a white man bc the white man is a threat then no one cares but if a police man kills a black man bc a black man is a threat everyone protests. I thought we were = but then black people are out there saying there lives matter. They do but white lives matter too so do blue lives and green lives and purple lives. Everyone matters but if you say that apparently you get in trouble. All lives matter but instead people had to make a program titled black lives matter.

Also guys you can love a guy as a friend just not in a gay way. That is not okay you are getting confused I love my friend who is the same gender as me but I don't get my feelings confused. Can yall just be straight please.

Step-by-step explanation:

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7/28 in reduced fraction

Bess [88]

7/28
(Now we divide both sides by 7 since they can easily be divided by 7)
7÷7=1
28÷7=4

Since we can't divide 1/4 anymore, 7/28 as a simplified fraction is 1/4.

3 0

3 years ago

Read 2 more answers

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