IGCSE MATHS Doubts

[princess12]

thanx
can any 1 explain me oct 2007 q 6,7
urgently fast
best of luck

[Ghost Of Highbury]

@princess - the whole question?? Q6 and Q7 ?? any specific subpart??

[Adzel]

o/n 2007 p4 q3 (explaination please)

Thanks in advance...

[princess12]

the whole question a@di please
 

[Ghost Of Highbury]

ans 6) ai) 1cm:200cm

               1cm:2m

              therefore AB = 26m/2m = 13 cm

                           BD = 30m/2m = 15 cm

                           Angle A shud be 80 degree.

aii) Measure them with ur protractor.

aiii) Bisect the angle DAB and extend it till it intersects the trapezium.

aiv) Draw the perpendicular bisector of AD.

av) check the attached diagram.

bi) Sine rule. (Sin D)/26 = (Sin 80)/30

Sin D/26 = 0.032826925

sin D = 26*0.032826925 = 0.8535

D = sin^-1 0.8535 = 58.6 degree ~ 59 degree.

bii) Angle BDC = 100-58.6 = 41.4

     (BC)^2 = 18² + 30² - 2*18*30*cos 41.4 = 1224 - 810.11 = 414

     BC = = 20.3m

biii) area of triangle ADB + BDC = 0.5 * 26 *30sin'41.4' + 0.5 *18* 30sin'41.4' = 436

will answer the 7th question soon

[princess12]

thanxs
what about q 7

[Dark Prince]

Can some1 tell me how to do part (e) of this Qs....(attached below)....tnkx

[Ghost Of Highbury]

o/n 2007 p4 q3 (explaination please)

Thanks in advance...

Ans 3ai) Point C --> Its x-co-ordinate is 0 and it lies in the curve y = x^2 + 1.

thus, just substitute the value x=0 in the equation --> y = 0^2 + 1 = 1

C --> (0,1)

3aii) with y-axis --> y = 4 - x.

any point in y-axis has the co-ordinate x=0.

apply x=0 in y = 4-x --> y = 4-0

thus, with y-axis --> (0,4)

with x-axis --> y = 4-x

any point in x-axis has y-co-ordinate = 0

apply y=0 in y = 4-x

0 = 4-x

x = 4

point --> (4,0)

b) y = mx + c , here m = gradient and c is y intercept.

write y = 4-x in this form

y = -x + 4    or     y = -1(x) + 4

here m = -1

gradient = -1

c) this means find the x-values for which the point on the curve has a negative gradient. in this, a parabola, all points in the part of the curve in the second quadrant has a negative gradient. Thus, when x<0, gradient = negative.

d) the two graphs are y = x^2 + 1     and   y = 4-x

meaning , x^2 + 1 = 4-x

x^2 + 1 -4 +x = 0

x^2 +x - 3 = 0

------------------

e) apply x = +/-

u shud get 1.3 and -2.3

 The answers i.e 1.3 and -2.3 are the x-cordinates of the point B and A respectively.

y = 4-x

y = 4-1.3 = 2.7

y = 4-x

y = 4+2.3 = 6.3

A - (-2.3, 6.3)         B - (1.3, 2.7)

(x1+x2)/2 = (-2.3+1.3)/2 = -0.5

(y1+y2)/2 = (6.3+2.7)/2 = 4.5

midpoint = (-0.5,4.5)

[the_grim_reaper]

for these kind of ques u hav to use some common sense .. even im kinda bad at sequences =\ unfortunately there are chances of this topic comin up .. because it didnt come in p2

i got this one for my mock test and i solved it by trial n error .. one thing is good tht they already mention the ans. so i try everything to get the ans

   n (n+1)
= -------   + (n + 1)
      2
=n2 + 3n + 2
--------------
      2
ans is not (n +1 )2 so i tried adding again

=n2 + 3n + 2                      n (n+1)   
-----------------        +    ---------
         2                                 2

=(n +1 )2

got it .. sorry i couldnt give u a proper explanation =\

Let me take a shot at this.
All right, in the previous part it is stated that the formula for the nth term is n(n+1)/k where we've found k to be 2.

So,  if n is n LOL then its n(n+1)/2 and if n is (n+1) so we have (n+1)((n+1)+1)/2 (replace n by n+1 since that is the nth term here)..

Next just add them
   n(n+1)     (n+1)(n+2)
= ------- + -----------
       2           2
= (n^2+n) + (n^2+2n+n+2)
 --------------------------
                  2
= n^2+n^2+n+2n+n+2
----------------------
              2
= 2n^2+4n+2
-------------
          2
Taking 2 as common, you have =2(n^2+2n +1)
                                             ------------
                                                     2
After simplification, you have n^2+ 2n +1 which is simplified becomes (n+1)(n+1) which is equal to (n+1)^2     


As far sequences are concerned, we all suck at em' mate. It's a universal issue .

[Adzel]

Ans 3ai) Point C --> Its x-co-ordinate is 0 and it lies in the curve y = x^2 + 1.

thus, just substitute the value x=0 in the equation --> y = 0^2 + 1 = 1

C --> (0,1)

3aii) with y-axis --> y = 4 - x.

any point in y-axis has the co-ordinate x=0.

apply x=0 in y = 4-x --> y = 4-0

thus, with y-axis --> (0,4)

with x-axis --> y = 4-x

any point in x-axis has y-co-ordinate = 0

apply y=0 in y = 4-x

0 = 4-x

x = 4

point --> (4,0)

b) y = mx + c , here m = gradient and c is y intercept.

write y = 4-x in this form

y = -x + 4    or     y = -1(x) + 4

here m = -1

gradient = -1

c) this means find the x-values for which the point on the curve has a negative gradient. in this, a parabola, all points in the part of the curve in the second quadrant has a negative gradient. Thus, when x<0, gradient = negative.

d) the two graphs are y = x^2 + 1     and   y = 4-x

meaning , x^2 + 1 = 4-x

x^2 + 1 -4 +x = 0

x^2 +x - 3 = 0

------------------

e) apply x = +/-

u shud get 1.3 and -2.3

 The answers i.e 1.3 and -2.3 are the x-cordinates of the point B and A respectively.

y = 4-x

y = 4-1.3 = 2.7

y = 4-x

y = 4+2.3 = 6.3

A - (-2.3, 6.3)         B - (1.3, 2.7)

(x1+x2)/2 = (-2.3+1.3)/2 = -0.5

(y1+y2)/2 = (6.3+2.7)/2 = 4.5

midpoint = (-0.5,4.5)

Thanx

[Ghost Of Highbury]

thanxs
what about q 7

typing it, will take 5 minutes.

[Ghost Of Highbury]

Ans 7) Check the diagram.

         cii) A transformation matrix. Reflection:

1. reflection in ml y=x it's (0 1)
                                   (1 0)

https://studentforums.biz/index.php/topic,1432.15.html   (6th post)

di) arrange 2 sets of co-ordinates of the triangle veritcally to form a matri

(8  10)  * (0.5 0)   = (4 5)
(6  12)     (0 0.5)     (3  6)

do the same with (6,10) and ull get the matrix (3)
                                                               (5)

so the new co-ordinates are
(8,6) --> (4,3)
(10,12)--> (5,6)
(6,10) --> (3,5)

plot them.

if u observe carefully, ull see dii) it is an enlargement of factor 0.5 and the center (0,0)

e) 0.5*10 = 5

    0.5 *  8 = 4
     0.5* 6 = 3

((5,12) (4,6)(3,10))

[Vin]

hey can any1 help me wid  nov 2000 Q 7(a)

a)   Translation ( - 3, -4)

b)   tan x = 4.6/1.4
x = tan^-1 73.1

c)   i) height of larger / height of smaller = 7/5 = 1.4

ii)look for the blue lines .. meet at x so x = 10
(10, 0)

d)   i) A = ½ * 4.8 * 1.4
    = 3.36cm²

ii) very tricky !!
for an equation in terms of y=x
1.4/4.8
=0.29
y=0.29x
meaning x is 0.29 times y


e)k² = 64  ..we are dealing with area
k = 8

f)hypo of T z²= 1.42²+4.8²
=5
hypo of similar triangle = k * 5
= 8 * 5

[Ghost Of Highbury]

Thanx

ur welcome.

[Dark Prince]

Can some1 tell me how to do part (e) of this Qs....(attached below)....tnkx

can some1 now do my Q....which i ve posted half an hour bfore.......if an1 knws how to do it...lol