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sp2606 [1]
4 years ago
12

There are 2 1/2 pizzas if each gets an eighth of a pizza how many people get a slice

Mathematics
1 answer:
Butoxors [25]4 years ago
8 0

Answer:

20.

Step-by-step explanation:

That would be 2 1/2 divided by 1/8

= 5/2 / 1/8

= 5/2 * 8

= 40/2

= 20.

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480 x 4 = 1920 x 3 = 5760 x 5 = 28800

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What is 16 : 24 in simplest form
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2/3

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Using this table below, what is the rate of change? Don't forget to include your units.
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(24,18)(28,21)
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3 0
3 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                    

8 0
4 years ago
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