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What is the solution to this equation?

Komok [63]

Answer:

B. x=75

Step-by-step explanation:

First, write out the equation as you have it:

x/5=15

Then, multiply both sides of the equation by 5/1:

5/1(x/5)=15(5/1)

Your result is:

x=75

5 0

3 years ago

Which answer choice is correct

maria [59]

Answer: c

Step-by-step explanation:

8 0

3 years ago

Experimental vs. Theoretical Probability:

liubo4ka [24]

Hello Nitagomez,

To start the numbers 1 through 6 are on a number cube.
Half of them are even, 2, 4, and 6.

So it makes sense that when you roll the cube 60 times, you should have 30 even numbers. The theoretical probability is 30.

A predication of 20 is 10 less than the expected value of 30.
10 out of 30 is 33%.

The predication is off by 33%, I would argue that it is not a very good prediction.

I hope this helps, Good luck!
MrEquation

6 0

3 years ago

What is 5 1/4 as a single fraction greater than 1

Sauron [17]

Answer:

21/4

Step-by-step explanation:

(5 * 4) + 1

20 + 1

21/4

7 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