maths help (2 Viewers)

[tacogym27101990]

Re: Immediate Help!!!!!!!!!11

please can sum1 solve my confusion by solving this?

the distance of a point from the line x = -3 is two fifths of its distance from the line y= -1. find the locus of the line. please show appropriate working. thanks.

is it meant to be y=-3??

 

[tommykins]

Let the P(x,y)
sqrt[ (x+3)² + (y-y)² ] = 2/5[ (x-x)² + (y+1)² ]
(x+3)² = 4/25(y+1)²

25(x+3)² = 4(y+1)²
25x² + 150x + 225 = 4y² + 8y + 4

25x² + 150x + 225 - 4y² - 8y - 4 = 0 is your locus.

 

[tommykins]

Re: Immediate Help!!!!!!!!!11

Let the P(x,y)
sqrt[ (x+3)² + (y-y)² ] = 2/5[ (x-x)² + (y+1)² ]
(x+3)² = 4/25(y+1)²

25(x+3)² = 4(y+1)²
25x² + 150x + 225 = 4y² + 8y + 4

25x² + 150x + 225 - 4y² - 8y - 4 = 0 is your locus.

 

[bubblesss]

Re: Immediate Help!!!!!!!!!11

Let the P(x,y)
sqrt[ (x+3)² + (y-y)² ] = 2/5[ (x-x)² + (y+1)² ]
(x+3)² = 4/25(y+1)²

25(x+3)² = 4(y+1)²
25x² + 150x + 225 = 4y² + 8y + 4

25x² + 150x + 225 - 4y² - 8y - 4 = 0 is your locus.

y r u squaring it? the answer is in x and y terms.

 

[tommykins]

回复: Re: Immediate Help!!!!!!!!!11

y r u squaring it? the answer is in x and y terms.

To get rid of the sqrt.

The answer is the same anyhow.

 

[tacogym27101990]

Re: Immediate Help!!!!!!!!!11

use the perpendicular distance formula

|x+3| = 2/5|y+1|
5x+15 =2y+2 and 5x+15=-2y-2
5x-2y+13=0 and 5x+2y+17=0

and i guess you'd have to exclude the point (-3,-1)

 

[bubblesss]

Re: Immediate Help!!!!!!!!!11

use the perpendicular distance formula

|x+3| = 2/5|y+1|
5x+15 =2y+2 and 5x+15=-2y-2
5x-2y+13=0 and 5x+2y+17=0

and i guess you'd have to exclude the point (-3,-1)

omg please teach me how to do it ur way???????? and also when to square the locus and when to not?

 

[bubblesss]

Re: Immediate Help!!!!!!!!!11

omg i have my examz tomoro some one please tell me when u do PA^2 =PB^2 and when u dont???????//

 

[bubblesss]

Re: Immediate Help!!!!!!!!!11

bumpety bump!!!!!!1111:uhhuh:

 

[tommykins]

回复: Re: Immediate Help!!!!!!!!!11

omg i have my examz tomoro some one please tell me when u do PA^2 =PB^2 and when u dont???????//

My answer and his answer is exactly the same.

You're better off grabbing some sleep, no point studying and then not having energy to do the exam properly

 

[bubblesss]

Re: 回复: Re: Immediate Help!!!!!!!!!11

My answer and his answer is exactly the same.

You're better off grabbing some sleep, no point studying and then not having energy to do the exam properly

how r the two answers the same?

 

[tommykins]

回复: Re: 回复: Re: Immediate Help!!!!!!!!!11

how r the two answers the same?

I graphed it. They're the same.

 

[-tal-]

Re: ASAP help!!!!!!!!!!!!!!!!!!!!!!!!!!11

q2

2^10 = 1024 = approximately 10^3

2^1000 = (2^10)^100 = approximately (10^3)^100 = about 300 digits

Since 2^10 = 1024, the 24 that was taken off in the bold part needs to be multiplied by 100 also, adding 2 more numbers on then end.

300 + 2 = 302

 

[lyounamu]

Re: ASAP help!!!!!!!!!!!!!!!!!!!!!!!!!!11

q2

2^10 = 1024 = approximately 10^3

2^1000 = (2^10)^100 = approximately (10^3)^100 = about 300 digits

Since 2^10 = 1024, the 24 that was taken off in the bold part needs to be multiplied by 100 also, adding 2 more numbers on then end.

300 + 2 = 302

Your working out is not very logical. You don't just multiply it by 100.

 

[untouchablecuz]

Re: maths again

P (x,y)
x - y = 0

a = 1
b = -1
c =0

So we have D = |(ax + by + c)/sqrt(a²+b²)|

sqrt2 = |x - y/sqrt(1²+[-1]²)| = |x-y/sqrt2|

Square both sides to get rid of the abosolute value.

2 = (x-y)²/2
4 = (x-y)² is your locus.

You can't square to get rid of absolute value.

|x| = surd (x^2)

|x| =/= (x^2)

 

[lyounamu]

Re: maths again

You can't square to get rid of absolute value.

|x| = surd (x^2)

|x| =/= (x^2)

Actually you can.

AV(x) = SR(x^2)

Sqaure both sides:

x^2 = x^2

tommy is pro at maths, don't argue with him.

 

[untouchablecuz]

Re: maths again

tommy is pro at maths, don't argue with him.

APPEAL TO AUTHORITY!

Anyway, you can square it? I've always been taught different.

 

[lyounamu]

Re: maths again

APPEAL TO AUTHORITY!

Anyway, you can square it? I've always been taught different.

Yeah. Because AV(x) = SR(x^2) but in here SR(x^2) =/= x

 

[untouchablecuz]

Re: maths again

Yeah. Because AV(x) = SR(x^2) but in here SR(x^2) =/= x

http://en.wikipedia.org/wiki/Absolute_value#Real_numbers

Wikipedia says different for the real number system and the complex number system (correct me if i'm wrong, cause I got no idea about complex numbers).

 

[lyounamu]

Re: maths again

http://en.wikipedia.org/wiki/Absolute_value#Real_numbers

Wikipedia says different for the real number system and the complex number system (correct me if i'm wrong, cause I got no idea about complex numbers).

Yeah, it's different depending on the different system you are talking about.

In the real number system, you cannot have negative root (e.g. SR(-4))

But in the complex number system, you can have one.

Here, I am talking in the context of the real number system...hope that cleared some up.