Math Questions some easy some hard (1 Viewer)

[Fortian09]

This question is probably easy but i cant seem to figure it out...

Find the H.C.F. of (2x-1)(x^2-6x+9) and (x^2-3x)(4x^2-1).

Simplify (2^(n+1)-2^(n-1))/(2^(n+1)+2^(n-1))

In triangleABC, if AB:BC:CA = 4:5:6, find cos A (no diagram)

0<SUP>0 </SUP>< x < 360. (sin x+1)(2cos x + 1)= 0, find the values of x.

thanks for the help more questions coming

 

[lyounamu]

Find the H.C.F. of (2x-1)(x^2-6x+9) and (x^2-3x)(4x^2-1).

(2x-1)(x^2 - 6x + 9) = (2x-1)(x-3)^2

AND (x^2-3x)(4x^2-1) = x(x-3)(2x+1)(2x-1)

Therefore, the H.C.F. is (x-3)(2x-1)

Simplify (2^(n+1)-2^(n-1))/(2^(n+1)+2^(n-1))

2^(n+1) - 2^(n-1) = 2. 2^n - 2^n/2 = 1.5 . n^2
2^(n+1) + 2^(n-1) = 2.2^n + 2^n/2 = 2.5 . n^2
Therefore, (2^(n+1)-2^(n-1))/(2^(n+1)+2^(n-1)) = (1.5 . n^2 )/ (2.5 . n^2 )= 3/5

In triangleABC, if AB:BC:CA = 4:5:6, find cos A (no diagram)

Let sides be equal to 4, 5 and 6.
Cos A = (4^2 + 6^2 - 5^2) / (2 . 4 . 6) = 9/16

Bonus:
A = 55.7711336 degrees
= 55 degrees 46 minutes

0<SUP>0 </SUP>< x < 360. (sin x+1)(2cos x + 1)= 0, find the values of x.

thanks for the help more questions coming

(sinx+1)(2cosx+1) = 0
sin x = -1 or cosx = -1/2
x= 270 degrees or x = 120 degrees or x = 240 degrees

 

[doink]

1. Factorise both equations, first one becomes:
(x-3)^2(2x-1). Second one:
[x(x-3)][2x-1][2x+1]

so HCF between them is (2x-1)(x-3).

2. See above

3. changing around cos rule gives:
CosA = b^2 + c^2 - a^2 / 2bc
sub in values for a, b, c

4. either one or both of the factors has to be equal zero for eqn to hold true.
therefore sinx = -1 and/or cosx= -1/2

find values and write them down for 0 - 360

 

[lyounamu]

1. Factorise both equations, first one becomes:
(x-3)^2(2x-1). Second one:
[x(x-3)][4x^2 -1]

so HCF between them is x-3.

2. Same base so you can work on exponentials
n+1 - n +1 / n+1 +n -1
2/2n
1/n

3. changing around cos rule gives:
CosA = b^2 + c^2 - a^2 / 2bc
sub in values for a, b, c

4. either one or both of the factors has to be equal zero for eqn to hold true.
therefore sinx = -1 and/or cosx= -1/2

find values and write them down for 0 - 360

I think you need to try Q1 & Q2 in more depth.

For Q2, you cannot use that method. According to the index Law, you can only do that when you have the algebra multiplying each other. In that case, it's a substraction and addition. You cannot use that method.

 

[minijumbuk]

*Confirms lyounamu's works*
*Reliability increases*

 

[lyounamu]

*Confirms lyounamu's works*
*Reliability increases*

Lol