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The angle of depression equals the angle of elevation because they are alternate angles on parallel lines.

The height is opposite the angle of elevation.  the horizontal length is adjacent.

You have the opposite, you want the adjacent so you would use     $$tan\theta=\frac{opp}{adj}$$

So we'd solve this like

10 + 6z - 3 + z - 5z  ( now we will bring the like terms together)

=10 - 3  + 6z + z - 5z   ( now simply add or subtract them as they r arranged0

=7 + 2z  ANSWER

i hope uve understood by now but if not them pls be free to ask!

i hope this helps!

Do you mean

$$125\sqrt[3]{a^6}-\sqrt[3]{125a^6}\\\\
=125\sqrt[3]{(a^2)^3}-\sqrt[3]{5^3(a^2)^3}\\\\
=125a^2-5*a^2\\\\
=120a^2$$

Answered here

Could you please tell me what you used to graph that Chris?

So we'd solve this lik:

6x + 5 = 20  ( now we will take 5 to the other side to subtract from 20)

6x = 20 - 5 ( now we'll simply just subtract it )

6x = 15  ( now we will take 6 to the other side and divide it by 15 which will then give us our x)

x = 15/6

x = 2.5 ANSWER

i hope u've understood by now but if not then pls be free to ask!

i hope this helps!

Zegroes that is only your thinking , that some people dont know what is 2+3 ,!but according to me and the truth 100%(or 99)of the world knows whats 2+3, i think u might sometimes get  confused with whats 2+3 , 5 or 4!Thats why yor saying " some people dont know what 2+3 is?"!

And Melody thats correct , these kind of answers like whats 2+3 shouldnt get points because they dont worth it!and 3 points is no less!

$${\sqrt{{\mathtt{3}}}} = {\mathtt{1.732\: \!050\: \!807\: \!568\: \!877\: \!3}}$$

$${\left|{\mathtt{1}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}\right|} = {\mathtt{0.732\: \!050\: \!807\: \!568\: \!877\: \!3}}$$

$${\left|{\mathtt{2}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}\right|} = {\mathtt{0.267\: \!949\: \!192\: \!431\: \!122\: \!7}}$$

Add up the last two (which are only approximate, as √3 is irrational) to get 1.

Or:  |1 - √3| = √3 - 1 because √3 is bigger than 1

       |2 - √3| = 2 - √3 because 2 is bigger than √3

Add these up and the √3's cancel out, leaving 2 - 1 = 1.

You can use the calculator on the Home page here to do this.

$${{\mathtt{4}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{\mathtt{7}} = {\mathtt{112}}$$

If you mean rotate a vector by 45° then multiply it by (1/2)i, where i = √(-1) to get a 45° anti-clockwise rotation.

Oops!  No, this will give a rotation by 90° and halve the magnitude!  You need to multiply by exp(i*pi/4).

"collinear" means along the same line; or if it refers to points, then it means the points lie along a straight line.

If you sum vectors A and B you get the same result as if you sum vectors B and A.

The picture with the chains means that the forces, F1 and F2, in two of the chains are added (vectorially) to give a resultant vector (F1+F2).  This is then added (vectorially) to the force F3 to give an overall resultant force FR.

Adding vectorially means you add the x-component values together, and add the y-component values together.

In the s***w eye you need to resolve F1 into it's components (F1*cos(theta1), F1sin(theta1)) and F2 into its components (F2cos(theta2), F2sin(theta2)), (where the thetas are angles between the vectors and the x-axis) and get the components of the resultant from

(FR_x, FR_y) = (F1*cos(theta1)+F2cos(theta2), F1sin(theta1)+F2sin(theta2))

The magnitude of the resultant is given by √( FR_x² + FR_y²) and its angle to the x-axis by tan^-1(FR_y/FR_x)

To make x on its own by doing what you do on one side to both sides of the equals sign. In this case to remove the division of 5 you would multiply and cancle it on the left hand side, then take that to the right hand side as multiplication. There for you have x = 5 * (-6)


To correct the last post it wouldn't be x and x = it would be 1 * x, which to make x on it's own you would still have to take the 1 to the other side as a division. Just imagine it is -6/4 instead and c4ross multiply would give you not x=.

$$\int 2x+1\;dx = \frac{2x^2}{2}+x+c\;=x^2+x+c$$

1. 
   
   
   

   Finite set is a  set that has a  finitenumber of elements. For example,

   
   

   in alphabets we have only 26 letters.

   
   

   {a,b,c,d,...................z}

   
   

    

   
   

   Infinite sets may be countable or uncountable.

   
   

   Some examples are:

   
   

   the  set of all integers, {..., -1, 0, 1, 2, ...}, is a countably  infinite set;

   
   

   the  set of all real numbers is an uncountably  infinite set.

   
   
   
   

Is F_R in the below image "resultant force"? 
Can some one please verbally interprut the image with more classification.

Image source Victoria University, Collaborate, Engineering Fundementals, 24/07/14

How is the bellow question in the image worked out? 

Image source Victoria University, Collaborate, Engineering Fundementals, 24/07/14

it = 2 okey 

a negative(-) and a negative(-) = a positive(+)

-,-=+

+,+=+

-,+=-

so its 3+7

which equals 10

I feel like you should be rewarded for answering even the most trivial questions i mean you never know some people dont know what 2+3 is?

CPhill has a slightly different attitude to me.  That is ok.  

We get on really well and it is good that we have differerent attitudes sometimes.  

If people all thought the same about everything the world would be very boring!

I love to watch the fun that goes on between CPhill and his 'rat pack'.

...What she said.....

I'm assuming this might be....

(x-2)/4 + 12/x = x/2   ????

Multiply though by 4 and we have

x - 2 + 48/x  = 2x     

Multiply through by x

x^2 - 2x + 48  = 2x^2

Rearranging, we have

x^2 + 2x - 48 = 0

Factoring produces

(x + 8) (x - 6) = 0

Set each factor to 0, and we have that.....  x = 6  and x = -8

I'll leave it for you to check that both solutions "work".....

I usually only give points for answers with good explanations.  

That doesn't always mean that they have to be correct but it does mean that you have attempted a question that you serioulsy thought you could answer and that you have explained your steps.

If a person gives a correct answer but with no explanatin I don't usually give points for that because it does not help anyone to learn anything.

I would not give points for questions such as 2+3 because these are not serious questions. (they are 'snark' questions) If people want to play with them that is fine.  But there should be no points involved.

I really enjoy watching the fun threads, I would never want to stop people from playing on the forum but I do not believe that people should get any points for it unless it is of exceptional quality.  For instance,  Rosala puts a lot of effort into her 'wraps' and they entertain a lot of people so i would give her points for that but I would not usually give any points for all the fun that follows on that thread.

I have a vision for this forum.  I want it to be a fun place to be, but i want the people who give proper mathematical answers to be given the credit that they deserve.  This is first and foremost a place of learning.  Learning should be fun but the credit (points) should mainly go to the serious teachers.  

This is my opinion.

MELODY ♬