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Mathematics: Post your doubts here!

bamteck

bamteck

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whitecorp said:
(iii) Notice that f(x) lies between 4-3=1 and 4+3=7 inclusive. Solving for f(x)=k can be interpreted as drawing a horizontal like y=k and finding the number of intersections with the y=f(x) curve. Hence, if no solutions arise, there are no intersections, ie k>7 or k<1.

(iv) A function has an inverse if and only if it is either strictly increasing or decreasing. Try sketching out the graph, you should be able to see that maximum value of A= pi radians, because in between pi/2 and pi radians, the function is strictly increasing. Alternatively, you can find dy/dx= -3 cosx, which is > or equals to zero for the domain defined.

(v) Let y=4-3sinx, then making x the subject gives x= arcsin (4-y)/3
Doing the reflection in the y=x line gives inverse of g to be = arcsin (4-x)/3
Hence, when x=3, g^-1 (3) =arcsin (4-3)/3 =arcsin (1/3).

Hope this helps. Peace.
Click to expand...

OMG ! You are great ! Thanks a lot :)
 
VelaneDeBeaute

VelaneDeBeaute

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iKhaled said:
thanks a lot. this constant thingy annoyed me a lot. sometimes they put it and sometimes they don't and i always used to wonder why sometimes they do and sometimes they dont!
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lol! Its always there but in limits, they just ignore it! :)
 
iKhaled

iKhaled

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Shahed Shubeir said:
Good afternoon ,
can anyone help me in question 7 part ii & iii may/june 2006
here's the link
it's urgent .
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s06_qp_4.pdf
Click to expand...
7 ii) they r moving on a smooth plane so a = g sinθ and the height is 1.6m and the distance moved is d = 2.6t so d = 2.6(2.5) which is 6.5m
now we need to find sinθ which is the height ( opposite ) / hypotenuse which is the distance moved so sinθ = 16/6.5
a = 10( 16/6.5)
a= 2.46 ms/2

7 iii) when q is at the highest point it means its final velocity is equal to zero so..
v = u + at
0 = -1.3 + 2.46t
t = 0.528s

now find the distance moved by particle p by substituting the t = 0.528 in the equation s = ut+ 1/2 at^2

i hope this helped u!
 
Taiyaba

Taiyaba

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Plz someone help me with this one :(
A line joining a vertex of a triangle to the mid-point of the opposite side is called a median. Find the length of the median AM in the triangle A(-1,1), B(0,3) and C(4,7).

Nd

A line has vertices A(-2,1), B(3,-4) and C(5,7).
(a) Find the coordinates of M, the midpoint of AB, and N midpoint of AC.
(b) Show that MN is parallel to BC

Can someone plz tel me hw to do part (b)

And

The points A(2,1), B(2,7) and C(-4,-1) form a triangle. M is the midpoint of AB and N is the midpoint of AC.
(a) Find the lengths of of MN and BC.
(b) Show that BC=2MN.
Only part (b)

Thnx in advance :)
 
VelaneDeBeaute

VelaneDeBeaute

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Taiyaba

Q.1. A line joining a vertex of a triangle to the mid-point of the opposite side is called a median. Find the length of the median AM in the triangle A(-1,1), B(0,3) and C(4,7). The situation depicts M as the midpoint of BC so first find that one. After you have the coordinates of the midpoint, go ahead applying the length/magnitude formula for coordinates of A and M to find te length of the median.

Q.2.A line has vertices A(-2,1), B(3,-4) and C(5,7).
(a) Find the coordinates of M, the midpoint of AB, and N midpoint of AC.
(b) Show that MN is parallel to BC
Can someone plz tel me hw to do part (b)
For parallel lines, the gradient is same, that is, m1=m2. So find the gradients of both MN and BC. If they're equal, then you've proven that the lines are parallel.

Q.3.The points A(2,1), B(2,7) and C(-4,-1) form a triangle. M is the midpoint of AB and N is the midpoint of AC.
(a) Find the lengths of of MN and BC.
(b) Show that BC=2MN.
Only part (b)
Find the lengths of both BC and MN using the length formula. The length of BC must be twice that of MN(e.g. if BC is 10, MN is 5).
 
Taiyaba

Taiyaba

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VelaneDeBeaute said:
Taiyaba

Q.1. A line joining a vertex of a triangle to the mid-point of the opposite side is called a median. Find the length of the median AM in the triangle A(-1,1), B(0,3) and C(4,7). The situation depicts M as the midpoint of BC so first find that one. After you have the coordinates of the midpoint, go ahead applying the length/magnitude formula for coordinates of A and M to find te length of the median.

Q.2.A line has vertices A(-2,1), B(3,-4) and C(5,7).
(a) Find the coordinates of M, the midpoint of AB, and N midpoint of AC.
(b) Show that MN is parallel to BC
Can someone plz tel me hw to do part (b)
For parallel lines, the gradient is same, that is, m1=m2. So find the gradients of both MN and BC. If they're equal, then you've proven that the lines are parallel.

Q.3.The points A(2,1), B(2,7) and C(-4,-1) form a triangle. M is the midpoint of AB and N is the midpoint of AC.
(a) Find the lengths of of MN and BC.
(b) Show that BC=2MN.
Only part (b)
Find the lengths of both BC and MN using the length formula. The length of BC must be twice that of MN(e.g. if BC is 10, MN is 5).
Click to expand...
Thnk u thnk u thnk u sooooooo mch :D :p
 
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