Post Your Problem Here (3 Viewers)

[laracroft]

hey i got a question, but i think it mite jsut be me being dumb, but wat the answer to this question...
find the area of the minor segment correct to 2 dec. plces given that the radius is 3.21 and the angle subtended is 1.22
ne help would be excellent, i rekon ive jsut done something dumb but newyas

 

[CM_Tutor]

Area of a minor segment is A = (r² / 2) * (@ - sin@), where @ is in radians.

 

[laracroft]

lol, i was rite that was a dumb question, coz i jsut had my calculator is the wrong mode lol my fault as ususal, thanks neways

 

[tradewind]

i got confused half way through. could you help me on this one.

find any points of inflexion in the curve y=2x^4 - 4x^3 - 24x^2 +6x-8

 

[jesshika]

find the second derivative and set it to zero ..
solve for x
den you test for changes in concavity

 

[speersy]

O.k an easy one for u maths buffs

Q) the sum of the radii of two circles is 100cm
a) If one of the circles has a radius of x cm, show that the sum of the areas of the two circles is given by
A= 2pi(x(squared) -40x + 400)

b)find the value of x for which A is least.


I know how to do b), however i cant prove a) for some weird reason maybe i don't know the formula for the area of a circle anymore

 

[CM_Tutor]

Originally posted by speersy
Q) the sum of the radii of two circles is 100cm
a) If one of the circles has a radius of x cm, show that the sum of the areas of the two circles is given by
A= 2pi(x(squared) -40x + 400)

If one circle has radius x cm, then the other has radius (100 - x) cm, and we know the area of a circle is pi * r²

So, A = pi * x² + pi * (100 - x)²
= pi * (x² + 100² - 2 * 100 * x + x²)
= pi * (2x² + 10000 - 200x)
= 2 * pi * (x² - 100x + 5000)

... which conclusively proves that the question is worng.

 

[CM_Tutor]

Originally posted by tradewind
i got confused half way through. could you help me on this one.

find any points of inflexion in the curve y=2x^4 - 4x^3 - 24x^2 +6x-8

Originally posted by jesshika
find the second derivative and set it to zero ..
solve for x
den you test for changes in concavity

So... y = 2x⁴ - 4x³ - 24x² + 6x - 8
dy/dx = 8x³ - 12x² - 48x + 6
d²y/dx² = 24x² - 24x - 48 = 24(x² - x - 2) = 24(x - 2)(x + 1)

For inflections, we examine d²y/dx² = 0, which has solutions x = -1 and x = 2.

When x = -1, y = 2(-1)⁴ - 4(-1)³ - 24(-1)² + 6(-1) - 8 = 2 + 4 - 24 - 6 - 8 = - 32 ----- Point is (-1, -32)
When x = 2, y = 2(2)⁴ - 4(2)³ - 24(2)² + 6(2) - 8 = 32 - 32 - 96 + 12 - 8 = - 92 ----- Point is (2, -92)

Now, we must test the nature of each of these, by checking if concavity changes sign.

x: -2 -1 0 2 3
sign of d²y/dx²: + 0 - 0 +

There is a change in concavity at both these points, and so (-1, -32) and (2, -92) are both points of inflexion.

 

[Zarathustra]

Originally posted by overwhelming
Prove that 1 = 2

let a and b each be equal to 1. Since a and b are equal

b^2=a.b (eq. 1)

since a equals itself it is obvious that
a^2=a^2

subtract eqn1 from eqn2

a^2 - b^2=a^2 - ab

factorise

(a + b)(a - b)= a(a-b)

Now divide both sides by (a - b)


which means

a + b=a

subtract 'a' from both sides

b=0

as b was set to one this means that 1=0


add 1 to both sides

2=1

:rofl:


nb I got this out of a book titled zero: biography of a dangerous idea, can anyone see anything wrong with this proof? How can 1=2 argggghhhh

 

[Heinz]

Originally posted by Zarathustra
let a and b each be equal to 1. Since a and b are equal

...

Now divide both sides by (a - b)

...

can anyone see anything wrong with this proof? How can 1=2 argggghhhh

division by 0

 

[Zarathustra]

Originally posted by Heinz
division by 0

Yes, you are right Heinz - have you seen this proof before, is it a common proof? This is the reason why you can't divide by zero, once you do that logic is destroyed - muhahaha...

 

[Heinz]

Originally posted by Zarathustra
Yes, you are right Heinz - have you seen this proof before, is it a common proof? This is the reason why you can't divide by zero, once you do that logic is destroyed - muhahaha...

yeah the "proofs" pretty common. and you really didnt have to add
a + b=a
subtract 'a' from both sides
b=0
as b was set to one this means that 1=0
add 1 to both sides
2=1

a+b=a is where it usually ends

 

[Xayma]

But if you want to get a+b=a to get 2=1
.
.
.
But a=b
.: 2a=a
.: 2=1

 

[speersy]

Originally posted by CM_Tutor

... which conclusively proves that the question is worng.

Yes the question must have been wrong, my teacher could not do it either. Don't you hate when that happens?

 

[Seraph]

Originally posted by KeypadSDM
Wait a second. This question's fundamentally incorrect.

Do you mean when the area's a minimum? Because the maximum area (assuming you have to cut it into 2 pieces) is just below 9 m^2

yea i meant a minimum when i was doing that question

 

[Seraph]

Btw if i have a bag containing 5 blue marbles , 4 red marbles and 2 yellow marbles and 1 green marble
WITHOUT ANY REPLACEMENTS

how do i fidn the probability that

Exactly two balls have been drawn

Without having to do p(bbb) + P(bbr) + P(bby) + p(bbg) + p(RBB) + P(YBB) + P(GBB)

 

[Calculon]

That question is ambiguous and strange

 

[stoydgen]

Hey how are ya'z? Anyways I'm stuck on this Trig Q, and the teacher was away today and we dont have him for the rest of week

"A circle has a chord Of 25mm with an angle of (pi/6) subtended at the centre. Find to 1 decimal place, the length of the Arc cut off by the chord."

And can someone please tell me what "subtended" actually means?

And by the way, any yr 12 students might wanna give this textbook a go, its called "maths in focus 2" by Margeret grove. Its changing my maths study habits

 

[Seraph]

Originally posted by Calculon
That question is ambiguous and strange

oops
Exactly Two BLUE balls have been drawn

 

[Winston]

Originally posted by stoydgen
Hey how are ya'z? Anyways I'm stuck on this Trig Q, and the teacher was away today and we dont have him for the rest of week

"A circle has a chord Of 25mm with an angle of (pi/6) subtended at the centre. Find to 1 decimal place, the length of the Arc cut off by the chord."

And can someone please tell me what "subtended" actually means?

And by the way, any yr 12 students might wanna give this textbook a go, its called "maths in focus 2" by Margeret grove. Its changing my maths study habits

Is this under Trignometric Functions?

If so... isn't there a formula for findin the length of the arc or something? something about theta = l/ r or something...