Mathematics: Post your doubts here!

[Rutzaba]

6 sin^2Q + 6 sinQ cosQ = sinQ cosQ - cos^2Q
6 sin^2Q + cos^2Q = -5 sinQ cos Q

(Divide by cos^2Q everywhere) : 6 tan^2Q + 1 = -5 tan Q

6 tan^2Q + 5 tanQ + 1 = 0
(3 tanQ + 1) (2 tanQ + 1) = 0
tan Q = (-1/3) or tan Q = (-1/2)

Hope It Helps

ohh stupid me ... i was closing sin square and cos square to form one. but that was stupid

now that means ive gone out of practice in 5 months only

 

[Yousif Mukkhtar]

One question:
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s11_qp_13.pdf
q10) iii) How do we know if g has an inverse?

 

[Just visiting]

Can anyone help me plz?
How to solve for that
Thanks in advance

 

[SalmanPakRocks]

One question:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_13.pdf
q10) iii) How do we know if g has an inverse?

Because it's a one to one function.

 

[Minato112]

Can anyone help me plz?
How to solve for that
Thanks in advance

log2 ( x^2 - x + 2) = 1 + 2 log2 x

1 can also be written as log 2 to base 2 (henceforth will be written as log 2.2)

log2 ( x^2 - x + 2) = log 2.2 + log2 x^2
log2 ( x^2 - x + 2) = log 2 (2x^2)

Cancel log 2 on both sides,

x^2 - x + 2 = 2x^2
x^2 + x - 2 = 0
(x + 2) (x - 1) = 0

x = -2 or x = 1

And finally since log cannot be negative, you eliminate -2 which leads to the only answer, x = 1

 

[Minato112]

Can anyone help me plz?
How to solve for that
Thanks in advance

You will find from g`(x) that it has no turning points. Thus we can conclude that it is one-one and therefore it has no inverse.

 

[Binyamine]

Can anyone help me plz?
How to solve for that
Thanks in advance

log ( x^ 2 - x + 2 ) = 1 + 2 log x

i am ignoring the base 2, since i cannot right subscript, hence i will make both the side have log such that we can ignore them on the following steps

log ( x^ 2 - x + 2 ) = log 2 + log ( x ^ 2 )

Recall log a + log b = log ( ab )

log ( x^ 2 - x + 2 ) = log ( 2x ^ 2 )

x^ 2 - x + 2 = 2x ^ 2

x^ 2 + x - 2 = 0
( x + 2 ) ( x -1 ) = 0

either x = -2 or x = 1

you should reject x=-2 since log cannot be negative. Therefore our answer is x =1 .

 

[Rutzaba]

Because it's a one to one function.

HAHAHAHAA tum sirf is hi kam k lie aya karo

 

[Rutzaba]

Because it's a one to one function.

i have seen u posting here three tyms... to say that same dialogue! xD epic ho gee ap

 

[Rutzaba]

Thanks @Ruztaba and Binyamine

You guys are live savers.

yar i cudnt even save my face you are talking bout life???

 

[sweetiepie]

Q. At What Simple Interest Rate An Amount triples itself in 6 years ?
A. 10% B. 20% C. 33 1/2 D. None Of These ?

Q.Find the compound interest of principal amount Rs 50000 at the rate 5% for 3 3/4 years?

Q. ax+by+c = 0 is the linear equation in variable -------------------

Q. Solve
(a) x-2[3x -2(x+1)+5]=16
(b) 4(3x -2)= 7 (2-5x)-5

Q. What is the interest on Rs 1880.90 for one year @ 5 1/2 % ?

Plz Answer Someone ???

 

[Rutzaba]

Q. At What Simple Interest Rate An Amount triples itself in 6 years ?
A. 10% B. 20% C. 33 1/2 D. None Of These ?

Q.Find the compound interest of principal amount Rs 50000 at the rate 5% for 3 3/4 years?

Q. ax+by+c = 0 is the linear equation in variable -------------------

Q. Solve
(a) x-2[3x -2(x+1)+5]=16
(b) 4(3x -2)= 7 (2-5x)-5

Q. What is the interest on Rs 1880.90 for one year @ 5 1/2 % ?

Plz Answer Someone ???

Q.3 wait
sweetiepie

x-2[ 3x-2x-2+5]=16
x-2[ x+3]=16
x-2x -6 =16
-x= 22
x=-22

b. 12x-8=14-35x-5
--------> 47x = 17

x= 7/47 i guess

 

[Minato112]

Are those questions from CIE?

 

[Rutzaba]

Are those questions from CIE?

a looser like me remembers how to do q 3 only

 

[sweetiepie]

Q.3 wait
sweetiepie

x-2[ 3x-2x-2+5]=16
x-2[ x+3]=16
x-2x -6 =16
-x= 21
x=-21

b. 12x-8=14-35x-5
--------> 47x = 17

x= 7/47 i guess

STop Making Fun of Mine I Need Right Answer Not The Guess

 

[Rutzaba]

STop Making Fun of Mine I Need Right Answer Not The Guess

whos making fun of you sweety? im making fun of my self... this is how to solve it

 

[sweetiepie]

whos making fun of you sweety? im making fun of my self... this is how to solve it

ITS WRONG

 

[Rutzaba]

ITS WRONG

HAWWWWWWWWW

 

[Rutzaba]

lemme c agn

 

[VampGirl]

Q11)
ii) replace f ' (x) with y : y = 6 (2x + 3)
2x+3 = 6/y
2x= 6/y - 3
x = (6/y - 3 ) / 2
replace x with f ' x and y with x ; so f ' (x) = (6/x - 3) / 2

Now to find the domain you need to visualize the graph of f ' (x). You can infer that x has a small range. Now, where does the graph of f ' (x) cut the x axis?
0 = (6/x - 3) /2
x= 2

and where does the graph meet y axis? You can assume that the graph meets y axis.
y = (6/x - 3x/x) / 2 (I have taken the LCM of the terms in the bracket)
y = (6/0 - 0/0) Note: Anything divided by zero is infinity. So the graph starts at (0 , infinity) and ends at ( 2, 0)

0<x<(or equal to) 2

Note that the x is greater than 0 , not greater than or equal to. It is because for any value of x greater than 0, the graph exists but on x=0 , y becomes infinity. I hope its kind of clear

Thnk u