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Mathematics: Post your doubts here!

S

sandman za

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RoOkaYya G said:
coz its their magnitude i.e distance
Click to expand...

i have vector AB (-2, -2, 4) which has magnitude 2*root(6)

vector AC (0, 3, 4) has magnitude 5

a.b = 0 -6 + 16 = 10

so we have 10 = 2*root(6) * 5 * cosBAC

cosBAC = root(6) / 6

cos inverse of that gives 65.9, which is not the same as cos inverse of 1/3 which is 70.5

have i made a mistake somewhere here?
 
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lxelle said:
HEYY.
The diagram shows a trapezium ABCD in which BA is parallel to CD. (ii) Given that the length of CD is 12 units, find the position vector of D. Diagram : http://onlineexamhelp.com/wp-content/uploads/2014/08/9709_s14_qp_12.pdf
Click to expand...
IMG_3472.jpg
 
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janasalem said:
October november 2011 paper 13 9709
can someone explain this ?

Functions f and g are defined by
f : x → 2x + 3 for x ≤ 0,
g : x → x^2 − 6x for x ≤ 3.

Find the set of values of x which satisfy gf(x) ≤ 16.
Click to expand...
f(x) and g(x) ARE related to each other by the function gf(x).

Solving (iii):
(2x+3)^2 - 6(2x+3)
= 4x^2 - 9 ≤ 16

= x^2 ≤ 25/4

x ≤ +/- 25/4

=> x ≤ +25/4 or x≤-25/4

The value of x must satisfy both the domains of g(x) and f(x)
Domain of g(x) is x must be less than or equal to 3, both +25/4 and -25/4 thus satisfy the domain of g(x)
However, the domain of f(x) is x must be less than or equal to 0. +25/4 obviously does not satisfy this domain and is therefore rejected.
=> -25/4 ≤ x ≤ 0
 
lxelle

lxelle

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Thought blocker said:
f(x) and g(x) ARE related to each other by the function gf(x).

Solving (iii):
(2x+3)^2 - 6(2x+3)
= 4x^2 - 9 ≤ 16

= x^2 ≤ 25/4

x ≤ +/- 25/4

=> x ≤ +25/4 or x≤-25/4

The value of x must satisfy both the domains of g(x) and f(x)
Domain of g(x) is x must be less than or equal to 3, both +25/4 and -25/4 thus satisfy the domain of g(x)
However, the domain of f(x) is x must be less than or equal to 0. +25/4 obviously does not satisfy this domain and is therefore rejected.
=> -25/4 ≤ x ≤ 0
Click to expand...
isn't the answer -5/2 ≤ x ≤ 0???
 
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RoOkaYya G said:
it wont work of course -___- coz u r searching for the angle A. which is between AB and AC....so u need to take A as starting point. its vectors so the direction counts hell lot. so u need to strt from A. so ull be needed vector directions AC and AB.thts how ive learnt it. ask others if theres other explanations.
KIDDO!! HOW DOES IT BECOME 4 WHEN U DO b-a FOR VECTOR AB? :eek: im getting -2 :(
for the i position ---> (4,-2,4)
7ii) use formula for area of triangle 0.5*AB*sinC
So in this case u shld AC and AB using distance formula
use the answer in part (i) and derive the sin BAC from it using identities. or u can even use right angled triangle n derive tht.
then u replace in area of triangle formula.
Click to expand...
Stuck in that green part. :( Rutzaba
 
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