Mathematics Marathon (1 Viewer)

[gurmies]

Is the probability 3/7 ?

If i'm correct: Prove that the parabolas y = 2x^2 - 6x + 7 and y = x^2 - 2x + 3 touch eachother and find the co-ordinates of the point of contact (Fitzpatrick 2 Unit)

 

[kurt.physics]

Is the probability 3/7 ?

I got this to.

 

[kurt.physics]

fail.

I smell a troll!!!

 

[bored of sc]

(1) y = 2x^2 - 6x + 7
(2) y = x^2 - 2x + 3

Hmm... Not sure how to prove it. They are both positive definite concave parabolas going off to infinity (and without any points of discontinuity) so it must conclude that they touch/cross at least once.
Domain is all real x for each one and range is y > 11.5 for (1) and y > 2 for (2)

Let (1) = (2)
2x²-6x+7 = x²-2x+3
x²-4x+4 = 0
(x-2)² = 0
x = 2

(1) y = 8-12+7 = 3
(2) y = 4-4+3 = 3

Point is (2,3)

 

[gurmies]

You would get most marks for that, but what the answer wanted was recognition of ∆ = 0 meaning only one solution...

 

[shady145]

ok my turn to ask one.
if a meal, including the tip of 15% was $178, what was the original price of the dinner. nice n easy 1 lol

 

[shady145]

o for that probability question i got 1/7
10/28+6/28-12/28=4/28=1/7

 

[gurmies]

115% =====> $178

100% =====> $x

x = (178 x 100)/115

= $154.78

Find the equation of the parabola with vertex (-2, 3), axis parallel to the y axis and passing through the point (-6, -5)

 

[bored.of.u]

115% =====> $178

100% =====> $x

x = (178 x 100)/115

= $154.78

Find the equation of the parabola with vertex (-2, 3), axis parallel to the y axis and passing through the point (-6, -5)

ok since x = -b/2a (axis of symetry)
.: -2 = -b/2a
.: -4a = -b
.: b = 4a
using the general form of a parabola y = ax^2 + bx + c
we sub in wat we worked out above to get:
y= ax^2 + 4ax + c
then using the two points specified we can make the two equations:
3= 4a-8a+c = -4a+c (subbing in the point -2,3) and ----> 1
-5= 36a-24a + c = 12a + c (subbing in the point -6,-5)--> 2
then we solve simutaneously
so let, 2-1
.: 12a + c = -5
- (-4a + c = 3)
.: 16a = -8
.: a = -1/2 or -0.5 <----sub into 1 to get:

2 + c = 3
c = 1
from before we know b = 4a so sub in 'a'
b = 4 x -0.5
b = -2

.: the equation of the parabola is y=-1/2x^2 - 2x + 1 OR y= -x^2 - 4x +2

hope this is correct i make a lot of silly errors. OK i will post a question now

Factorise fully: (a^2 - b^2 - c^2)^2 - 4b^2 c^2

 

[kaz1]

ok since x = -b/2a (axis of symetry)
.: -2 = -b/2a
.: -4a = -b
.: b = 4a
using the general form of a parabola y = ax^2 + bx + c
we sub in wat we worked out above to get:
y= ax^2 + 4ax + c
then using the two points specified we can make the two equations:
3= 4a-8a+c = -4a+c (subbing in the point -2,3) and ----> 1
-5= 36a-24a + c = 12a + c (subbing in the point -6,-5)--> 2
then we solve simutaneously
so let, 2-1
.: 12a + c = -5
- (-4a + c = 3)
.: 16a = -8
.: a = -1/2 or -0.5 <----sub into 1 to get:

2 + c = -3
c = -5
from before we know b = 4a so sub in 'a'
b = 4 x -0.5
b = -2

.: the equation of the parabola is y=-1/2x^2 - 2x -5 OR y= -x^2 - 4x -10

hope this is correct i make a lot of silly errors. OK i will post a question now

Factorise fully: (a^2 - b^2 - c^2)^2 - 4b^2 c^2

(a²-b²-c²+2bc)(a²-b²-c²-2bc)
[a²-(b²+c²-2bc)][a²-(b²+c²+2bc)]
[a²-(b-c)²][a²-(b+c)²]
(a-b+c)(a+b-c)(a-b-c)(a+b+c)

 

[gurmies]

I would have done this:

We know vertex is (-2, 3) and it's in the form x^2 = 4ay (axis of parabola parallel to y-axis)

So, (x+2)^2 = 4a(y-3)

Subbing in the point (-6, -5)

16 = 4a(-8)

-32a = 16

a = -1/2

(x+2)^2 = 4(-1/2)(y-3)

(x+2)^2 = -2(y-3)

For your question:

(a^2 - b^2 - c^2)^2 - 4b^2 c^2

(a^2 - b^2 - c^2 - 2bc)(a^2 - b^2 - c^2 + 2bc)

[a^2 -(b^2 + 2bc + c^2)][a^2 -(b^2 - 2bc + c^2)]

[a^2 - (b + c)^2][a^2 - (b - c)^2]

(a - b - c)(a + b + c)(a - b + c)(a + b - c)

Given Sn = 17n - 3n^2, find an expression for the nth term

 

[Nevermore]

Is the probability 3/7 ?

If i'm correct: Prove that the parabolas y = 2x^2 - 6x + 7 and y = x^2 - 2x + 3 touch eachother and find the co-ordinates of the point of contact (Fitzpatrick 2 Unit)

Yea it is [maths]3/7[/maths]

 

[bored.of.u]

Given Sn = 17n - 3n^2, find an expression for the nth term

For gurmies' question:

Sn = {n[2a + (n-1)d]}/2 AND Tn = a + (n-1)d

So rearranging Sn :
Sn = {n[a + a + (n-1)d]}/2
Sn = [n(a + Tn)]/2
2.Sn = n(a + Tn)
(2.Sn)/n = a + Tn
.'. Tn = [(2.Sn)/n] - a

Then substituting Sn = 17n - 3n^2
Tn = {[2(17n - 3n^2)]/n} - a
Tn = [2(17-3n)] - a
Tn = 34 - 6n - a

Is this correct its looks kinda wierd =.=

So my question(easy):

Find all values for x:

sin^4 (x) - cos^2 (x) = -1 for 0<x<360

 

[Nevermore]

Find the solution for [maths]4x<=15<=-9x[/maths]

 

[gurmies]

Incorrect, your method is far more complicated than what's required

 

[electrolysis]

 

[kurt.physics]

hence

 

[kurt.physics]

Therefore it's 27/3-8/3

19/3

whoops! I'll fix it up

 

[gigglinJess]

Seen there doesn't seem to be a question going.....

Find the volume of the solid formed when y = (x + 1)^2 is rotated about the x-axis from x = 0 to x = 4 (4 marks)

 

[AnandDNA]

hence

what do you use to be able to type your answer like that??