Maths, Addmaths and Statistics: Post your doubts here!

[bikermicefrmars]

AOA people. how to do questions like these:

Find the smallest positive integer, n, such that 168n is a square number.?

 

[scouserlfc]

its simple break 168 into its square factors and u will know it !
Im sure people here will help u in detail !

 

[bikermicefrmars]

Find the fraction which is exactly halfway between 5/9 and 8/9 .

How to solve this?

 

[Anon]

Take average of both the values..

5/9 + 8/9
...2

= 13/18


You can also mentally solve it as:

5/9 and 8/9

the mid value between 5 and 8 is 6.5,

therefore:
6.5/9
=13/18

 

[Anon]

AOA people. how to do questions like these:

Find the smallest positive integer, n, such that 168n is a square number.?

Its simple,
as scouser said, break it down to its factors...

168= 3x2^3x7
=2 x 2 x 2 x 3 x 7

now when we multiply it by n, the number becomes a square number,
it should be squared when you are able to pair up two factors:

(2 x 2 x 2 x 3 x 7) multiplies by (2 x 3 x 7)

n = (2 x 3 x 7)
n = 42

 

[Anon]

I have a little query in Add.Maths derivative and integration.

Here is the trouble:

IN a question, I first find derivative of y = x √(2x + 15)

The answer to it is: y= 3(x+5) / √(2x+15)


NOw in next part we have to integrate (x+5) / √(2x+15)
by comparing the previous answer...

Now what I want to know is that, will we add a "c" in the answer ?? It is indefinite integral, but on other hand, we arent solving, we are comparing...

Any ideas!?

 

[Nibz]

No matter what you're doing. Comparing, solving - anything. You have to put 'c' or 'k' there. Compare it and simply put 'c' in the end.

 

[Anon]

Are you sure ??

coz I didnt add in the previous class test.... sigh.. poor me

 

[Nibz]

Yes! It's a must for indefinite integrals.

 

[Anon]

Another ques:

When finding area, how do we find M?
I have found C as (14, -3), D and M remains...

 

[Anon]

Just after posting, I figured out how to solve the question...

 

[scouserlfc]

LOL u are some genius !

 

[Anon]

LOL u are some genius !

 

[Waleed007]

Another ques:

When finding area, how do we find M?
I have found C as (14, -3), D and M remains...

How did u solve it?

 

[Anon]

How I solved:

1. Find gradient of line AB.
2.Find m2 (m1 x m2 =-1)
3.Now you have the gradient of AC
4.Find equation of AC
5. Substitute x=14 in equation, you get coordinates of C

Now here was the part I was stuck earliar:

We know x coordinate of D is 14.
x coordinate of B = -2.

6. Find the x coordinate of Midpoint of BD (12/2 = 6)
7.Now you have x coordinate of M.
8.Put that x value in equation you made in 4.
9. You now have coordinates of M.

Now you know, the mid value of BD is M.
value of y coordinate of B= 10
value of y coordinate of M= (see from step 9)

So find the coordinates of D.


Now consider Triangle, BCD and ABD, and find their area.
Add both the areas...

You're done

 

[Waleed007]

can we use the y-y1=m(x-x1) to find the equation of straight line ?

 

[abcde]

can we use the y-y1=m(x-x1) to find the equation of straight line ?

Of course. It's faster, too. I mean you don't need to bother to calculate 'c'.

 

[Waleed007]

Here's my Question:
Find the values of k if the lines 2x-5=ky and (k+1)x=6y-3 have the same gradient.....

 

[abcde]

Here's my answer:
ky = 2x -5
=> y = 2/k x - 5/k
Gradient of first line = 2/k
Gradient of second line = (k+1)/6
Equate 'em: 2/k = (k+1)/6
Solve the resulting quadratic equation of k^2 + k -12= 0 to get k= 3 or k = - 4.

 

[Waleed007]

Thanks!

but I have a problem how did you get the gradient of first line and second line?