Mathematics: Post your doubts here!

[Rutzaba]

Yes definitely. Please remember me too. Got a school test tomorrow, and I am SOO blank
Thanks for the detailed explanation. Thanks!

i sure will

 

[Rexwong97]

Hey the formula arc length equals to radius times subtended angle allowed to be used in mathematics o level code 0580

 

[Dnt knw]

so hws the m2 exm yesterday guys

 

[Dnt knw]

so hws the m2 exm yesterday guys

 

[snowbrood]

wen two lines are perpendicularthe b1 or directional vector becomes the normal to the ohter vector so here we have n1= 12, 6, -6)
which we took out by (7, 7, -5) -( (-5,1,1) = 12, 6,-6
then we apply formula
r.n=a.n
where r= x y and z
and a is any point on plane
wch can also be A as it passes thru A (1,4,-2)
then we have (x,y,z) . (12, 6 ,-6)= (1, 4,-2). (12, 6. -6)
you will get 12x+6y-6z=48

do u know how to do decimal search using chord approximation
Remember me in ur prayers please

 

[Rutzaba]

which topic is this?

 

[biscuitbiscuit]

Guys can you solve this " In a triangle OAB, O is the origin, A is (0,6) and B is (6,0). Find the equations of the three medians of the triangle"

 

[Saad Mughal]

Guys can you solve this " In a triangle OAB, O is the origin, A is (0,6) and B is (6,0). Find the equations of the three medians of the triangle"

Solution:
Midpoint OA = (0,3)
Midpoint OB = (3,0)
Midpoint AB = (3,3)

For first median, O to AB:
Gradient = 1
y - 3 = 1 (x - 3)
y = x

For second median, A to OB:
Gradient = 6-0/0-3 = -2
y - 6 = -2 (x - 0)
y = -2x + 6

For third median, B to OA:
Gradient = 0-3/6-0 = -1/2
y - 0 = -1/2 (x - 6)
2y = -x + 6

 

[biscuitbiscuit]

Solution:
Midpoint OA = (0,3)
Midpoint OB = (3,0)
Midpoint AB = (3,3)

For first median, O to AB:
Gradient = 1
y - 3 = 1 (x - 3)
y = x

For second median, A to OB:
Gradient = 6-0/0-3 = -2
y - 6 = -2 (x - 0)
y = -2x + 6

For third median, B to OA:
Gradient = 0-3/6-0 = -1/2
y - 0 = -1/2 (x - 6)
2y = -x + 6

Thanks

 

[biscuitbiscuit]

"Two sides of a parallelogram are formed by parts of the line 2x -y = -9 and x - 2y = -9." I was able to find the find co-ordinates of the vertex where they intersect which is (-3,3). The next part says that another vertex of the parallelogram is the point (2,1) and I have to find the equations of the other two sides of the parallelogram. Someone please explain this

 

[biscuitbiscuit]

"Two sides of a parallelogram are formed by parts of the line 2x -y = -9 and x - 2y = -9." I was able to find the find co-ordinates of the vertex where they intersect which is (-3,3). The next part says that another vertex of the parallelogram is the point (2,1) and I have to find the equations of the other two sides of the parallelogram. Someone please explain this

 

[Rutzaba]

type the exact question from the book plz

 

[Saad Mughal]

"Two sides of a parallelogram are formed by parts of the line 2x -y = -9 and x - 2y = -9." I was able to find the find co-ordinates of the vertex where they intersect which is (-3,3). The next part says that another vertex of the parallelogram is the point (2,1) and I have to find the equations of the other two sides of the parallelogram. Someone please explain this

They have given you the vertex, which means that you know that (2,1) lies on both the lines for which you have to the find the equation.
Since it is a parallelogram, opposite sides will be PARALLEL, which means equal gradients for opposite side lines.

Solution:
2x - y = -9
Gradient = 2
(2,1) lies on line,
y - 1 = 2 (x - 2)
y = 2x - 3 --> One equation.

x - 2y = -9
Gradient = 1/2
(2,1) lies on line,
y - 1 = 1/2 (x - 2)
2y - 2 = x - 2
2y = x --> Second equation.

 

[biscuitbiscuit]

They have given you the vertex, which means that you know that (2,1) lies on both the lines for which you have to the find the equation.
Since it is a parallelogram, opposite sides will be PARALLEL, which means equal gradients for opposite side lines.

Solution:
2x - y = -9
Gradient = 2
(2,1) lies on line,
y - 1 = 2 (x - 2)
y = 2x - 1 --> One equation.

x - 2y = -9
Gradient = 1/2
(2,1) lies on line,
y - 1 = 1/2 (x - 2)
2y - 2 = x - 2
2y = x --> Second equation.

Thank you so much. Now can you tell me how will I find the co-ordinates of the other two vertices and btw the first equation should be '2x - y = 3'

 

[Saad Mughal]

Thank you so much. Now can you tell me how will I find the co-ordinates of the other two vertices and btw the first equation should be '2x - y = 3'

Yeah, sorry, I solved it in a hurry and hence the mistake.
For other two vertices, you need to simultaneously solve the NON parallel line equations.
So, you need to simultaneously solve:
(i) y = 2x -3 and x - 2y = -9
(ii) 2x - y = -9 and 2y = x

 

[biscuitbiscuit]

Yeah, sorry, I solved it in a hurry and hence the mistake.
For other two vertices, you need to simultaneously solve the NON parallel line equations.
So, you need to simultaneously solve:
(i) y = 2x -3 and x - 2y = -9
(ii) 2x - y = -9 and 2y = x

Thanks

 

[snowbrood]

Yeah, sorry, I solved it in a hurry and hence the mistake.
For other two vertices, you need to simultaneously solve the NON parallel line equations.
So, you need to simultaneously solve:
(i) y = 2x -3 and x - 2y = -9
(ii) 2x - y = -9 and 2y = x

calculate the value of integral (range 1 to 13) 1/(2x+1) giving your answer in an exact form simplified as far as possible help guys

 

[Saad Mughal]

calculate the value of integral (range 1 to 13) 1/(2x+1) giving your answer in an exact form simplified as far as possible help guys

Solution:
y = 1/(2x+1)
Integral = [ln(2x+1)]/2 + c
Using range 1 to 13,
Integral = { [ln(27)/2] - [ln(3)/2] } (We ignore the constant c in definite integrals).
Integral = (ln 27 - ln 3)/2
Integral = (ln 9)/2

 

[shazmina]

Assalamu alaikum ............i really wanna know wether its important to solve the old years papers for pure mathematics like in 90s...
or else solving from 2002 would be enough ............
please reply....
thanks in advance

 

[Saad Mughal]

Assalamu alaikum ............i really wanna know wether its important to solve the old years papers for pure mathematics like in 90s...
or else solving from 2002 would be enough ............
please reply....
thanks in advance

The types of questions that came in the 1990s don't come in papers now but they were harder than the current papers so if you want to solve them for practice then go ahead, it'll only make your Maths better.