Maths Game (1 Viewer)

[xXmuffin0manXx]

[quote=lyounamu]Would you like to elaborate on that? I don't understand what you mean.[/quote]

Might be wrong but:
log(base2)((x(x^2-4))/x+2))=3
8=x(x^2-4)/x+2
(factorise)
x(x-2)(x+2)/x+2=8
x^2-2x-8=0
factorise
x=4 or -2

is that rite?

 

[lyounamu]

[quote=lyounamu]Would you like to elaborate on that? I don't understand what you mean.

Might be wrong but:
log(base2)((x(x^2-4))/x+2))=3
8=x(x^2-4)/x+2
(factorise)
x(x-2)(x+2)/x+2=8
x^2-2x-8=0
factorise
x=4 or -2

is that rite?

[/quote]

Because he/she took the log(base 2) out.

 

[xXmuffin0manXx]

Might be wrong but:
log(base2)((x(x^2-4))/x+2))=3
8=x(x^2-4)/x+2
(factorise)
x(x-2)(x+2)/x+2=8
x^2-2x-8=0
factorise
x=4 or -2

is that rite?

Because he/she took the log(base 2) out.[/quote]

oh right right rght..

my bad

thanks

 

[studentcheese]

Prove that the following lines are concurrent:
3x + 4y - 14 = 0
2x - 3y + 19 = 0
3x - 2y + 16 = 0

 

[pwoh]

Prove that the following lines are concurrent:
3x + 4y - 14 = 0
2x - 3y + 19 = 0
3x - 2y + 16 = 0

I give up.

 

[suzlee]

Prove that the following lines are concurrent:
3x + 4y - 14 = 0
2x - 3y + 19 = 0
3x - 2y + 16 = 0

That means it has a common point that all the lines cross, right?

3x + 4y - 14 = 0 ........................(1)
2x - 3y + 19 = 0 ........................(2)
3x - 2y + 16 = 0 .........................(3)




Finding point of intersection of (1) and (2)

(1)x2
6x + 8y - 28 = 0 ...............(4)

(2)x3
6x - 9y + 57 = 0 .......................(5)


(4)-(5)
17y - 85 = 0
y = 5 ........................(6)


sub (6) -> (2)

2x-3(5)+19=0
2x-15+19=0
x=-2



so (-2,5)......





Seeing if (3) crosses (-2,5)



3(-2) - 2(5) + 16 = 0

-6 - 10 + 16 = 10 TRUE




therefore the three lines are concurrent

 

[pwoh]

That means it has a common point that all the lines cross, right?

3x + 4y - 14 = 0 ........................(1)
2x - 3y + 19 = 0 ........................(2)
3x - 2y + 16 = 0 .........................(3)




Finding point of intersection of (1) and (2)

(1)x2
6x + 8y - 28 = 0 ...............(4)

(2)x3
6x - 9y + 57 = 0 .......................(5)


(4)-(5)
17y - 85 = 0
y = 5 ........................(6)


sub (6) -> (2)

2x-3(5)+19=0
2x-15+19=0
x=-2



so (-2,5)......





Seeing if (3) crosses (-2,5)



3(-2) - 2(5) + 16 = 0

-6 - 10 + 16 = 10 TRUE




therefore the three lines are concurrent

:O oh, THAT's what it means.

Uh...solve:

6x^2 + 7x - 3 = 0

 

[suzlee]

Uh...solve:

6x^2 + 7x - 3 = 0

(6x+9)(6x-2)
---------------- = 0
6

= (2x+3)(3x-1) = 0

therefore x=-3/2, 1/3



Q
A sprinkler sprays water in an arc of 240 degrees with a radius of 4.5m. What is the area covered by the sprinkler?

 

[pwoh]

(6x+9)(6x-2)
---------------- = 0
6

= (2x+3)(3x-1) = 0

therefore x=-3/2, 1/3



Q
A sprinkler sprays water in an arc of 240 degrees with a radius of 4.5m. What is the area covered by the sprinkler?

240/360 x 4.5^2 x pi = 13.5pi
= 42.4 metres squared

Refer to the attached image.

KMNP is a square and triangle XMN is equilateral.
Prove that triangle KMX is congruent to triangle PNX.

 

[robibro]

Q
A sprinkler sprays water in an arc of 240 degrees with a radius of 4.5m. What is the area covered by the sprinkler?

lets see

A= 240/360 x pi x 4.5^2
= 42.4115...
= 42.4m^2 (to 1 dp)

 

[Nasonex]

Differentiate x^2

 

[gurmies]

answer to the log question is x = 4, no debate

y = x^2

y' = 2x

 

[Kabuzo01]

I'm to busy studying to even bother reading all this algebraness =P

 

[gurmies]

For congruence question, XNM is equilateral, i.e all angles are 60 degrees and all sides are equal.

Thus:

<PNX = < KMX = 30* (complementary angles) A

NM = PN = KM (NM is a side of a square, as are PN and KM) S

XM = XN ( sides of equilateral triangle) S

Therefore, triangle PNX is congruent to triangle KMX (SAS)

 

[suzlee]

240/360 x 4.5^2 x pi = 13.5pi
= 42.4 metres squared

Refer to the attached image.

KMNP is a square and triangle XMN is equilateral.
Prove that triangle KMX is congruent to triangle PNX.

assuming the point in the middle is X...

XN=XM (sides of equalateral triangle are =)
PN=KM (sides of square are =)

angle XNM = angle XMN (angles in equalateral tri. are =)
angle PNM = angle NMK (angles in square are 90)
therefore angle PNX= angle KMX (interior angles in square are all complementary??)

therefore tri. KMX is congruent to tri. PNX



I don't know if I reasoned them all correctly Can someone check it?





ohhhhhhh right ^

 

[Kabuzo01]

Wow nearly 1k views =P Looks like making this thread was useful lolz

 

[gurmies]

^^^That is correct =)

 

[Kabuzo01]

YAY 1k views.... but it looks like its dying off....

 

[Nasonex]

1) Rationalise the denominator:

1 - *root 2*
----------------
2 - *root 8*

 

[studentcheese]

1) Rationalise the denominator:

1 - *root 2*
----------------
2 - *root 8*

root 16
----------
8