Urgent Maths help needed (2 Viewers)

[marcquelle]

thank you all so much

: )

 

[marcquelle]

here is one that i did can you tell me if i have completed this:
2CosB-√3=0 for 0* <B<360*
2CosB-√3=0
2CosB=√3
CosB=√3/2
CosB=30*
CosB=360*-30*
CosB=30*
Therefore CosB=30*,330*

is that right please

 

[lyounamu]

here is one that i did can you tell me if i have completed this:
2CosB-√3=0 for 0* <B<360*
2CosB-√3=0
2CosB=√3
CosB=√3/2
CosB=30*
CosB=360*-30*
CosB=30*
Therefore CosB=30*,330*

is that right please

It's right but the working-out is not smooth enough.

I will just alter the last three lines, I would personally do:

cos B = SR(3)/2
B = 30, 360-30
= 30 or 330

 

[marcquelle]

ok thank you

 

[ratcher0071]

here is one that i did can you tell me if i have completed this:
2CosB-√3=0 for 0* <B<360*
2CosB-√3=0
2CosB=√3
CosB=√3/2
CosB=30*
CosB=360*-30*
CosB=30*
Therefore CosB=30*,330*

is that right please

After the Highlighted Point you are Cos^-1 both sides to get rid of the Cos, that's why lyounamu corrected.

 

[marcquelle]

"Find the equation of the normal to the curve y=2x^3-3x at the point P on itwhere x = -1" would that be
//m=dy/dx
=2x^3-3x
=6x^2-3
=6(-1)^2 -3
//m=3
//m=m2=-1/m1
//m=-1/3
y-y1=m(x-x1)
y-y1=-1/3(x+1)
y=2(-1)^3 -3(-1)
y=1
y-y1=m(x-x1)
y-1=-1/3(x+1)
3(y-1)=-1(x+1)
3y-3=-1x-1
x+3y-2=o

is that right?

sorry about all the questions but my test is tomorrow and our teacher/s never gave us an answer sheet so i have no way of knowing if I the material or not :S

 

[marcquelle]

yep saw that when he corrected, but thank you for clearing it up in case i didn't realise.

thank you all again i know i'm being a pain

 

[bubblesss]

"Find the equation of the normal to the curve y=2x^3-3x at the point P on itwhere x = -1" would that be
//m=dy/dx
=2x^3-3x
=6x^2-3
=6(-1)^2 -3
//m=3
//m=m2=-1/m1
//m=-1/3
y-y1=m(x-x1)
y-y1=-1/3(x+1)
y=2(-1)^3 -3(-1)
y=1
y-y1=m(x-x1)
y-1=-1/3(x+1)
3(y-1)=-1(x+1)
3y-3=-1x-1
x+3y-2=o

is that right?

sorry about all the questions but my test is tomorrow and our teacher/s never gave us an answer sheet so i have no way of knowing if I the material or not :S

yep yep that seems rite.

 

[Timothy.Siu]

for that question, u cud just find y by subbing in x in the original equation.
and i think thats right, ur method is right excerpt for taking unnecessary steps

u shud find the y value first and then differentiate to find the gradient of the normal then just point gradient formulae

 

[marcquelle]

i find if i don't do the unecessary steps i screw up so, i do them to be on the safe side. thanks and i have another so sorry about it but
"Differentiate with RESPECT to x" whats with "respect to" mean?
i) 5/x^4 - 5/4√x
this is what i got without knowing/remembering what "respect to" means.

d/dx=5x^-4 - 5^-1/4
=-20x^-5 - 5^-1/4

then thats it.

 

[ratcher0071]

i find if i don't do the unecessary steps i screw up so, i do them to be on the safe side. thanks and i have another so sorry about it but
"Differentiate with RESPECT to x" whats with "respect to" mean?
i) 5/x^4 - 5/4√x
this is what i got without knowing/remembering what "respect to" means.

d/dx=5x^-4 - 5^-1/4
=-20x^-5 - 5^-1/4

then thats it.

"With respect to x" means dy/dx

I think

 

[marcquelle]

oh thats simple duh i'm an idiot sometimes

i have another question again as well.
x=√6-3, express x-1/x in its simplest form with a rational denominator.

i have absolutely no idea, but i think it is going to be fairly obvious once someone says it though lol

thanks again guys

 

[bubblesss]

oh thats simple duh i'm an idiot sometimes

i have another question again as well.
x=√6-3, express x-1/x in its simplest form with a rational denominator.

i have absolutely no idea, but i think it is going to be fairly obvious once someone says it though lol

thanks again guys

x-1/x = sqr 6 - 3 - 1/sqr 6 - 3
= (sqr6-3)^2 - 1 / sqr 6 - 3
= (6 +9 - 6 sqr6 - 1)/ sqr 6 - 3
= 14 - 6 sqr 6/ sqr 6 - 3
= 2(7-3sqr6)/sqr 6 - 3
taking conjugate,
2(7-3sqr6)/sqr6-3 * sqr 6 +3/ sqr6 +3
= 2( 7 sqr6 +21-18-9sqr6)/ 6 - 9
=2 ( -2 sqr6 +3)/-3
= -4sqr6 +3/ -3
= (4sqr 6 - 3)/ 3

 

[marcquelle]

it wasn't obvious but i understand it now thanks so much

 

[Aplus]

Um...that doesn't really matter (unless the question specifically tells you to use that method). In many of the questions, you will sometimes find that this way of method is better. But if the question is hard to factorise and everything, it's best to go with what you suggested there.

I'm not sure. I just find that way easier because all you do is add and subtract fractions.

 

[lolokay]

i have another question again as well.
x=√6-3, express x-1/x in its simplest form with a rational denominator.

this one would be easier/less messy to solve if you evaluate 1/x first
1/(√6-3) * (√6+3)/(√6+3)
= (√6+3)/(6-3²
= -(√6+3)/3
x - 1/x
= (3√6-9)/3 - -(√6+3)/3
= (3√6 + √6 - 9 + 3)/3
= (4√6 - 6)/3
= 2(2√6 -3)/3 or 4/3 √6 - 2 [second looks simpler imo, but first is probably the simplest form. not sure if it matters]

+ how do you get the square root sign?

 

[marcquelle]

thanks again guys i actually have done the exam i think i have done better then i have in past texts

 

[Pink Oni]

thanks again guys i actually have done the exam i think i have done better then i have in past texts

LOL, hey marcquell, Extension English is getting to you I think. It happens to me too xD

I feel the same about maths, high five!

 

[marcquelle]

LOL, hey marcquell, Extension English is getting to you I think. It happens to me too xD

I feel the same about maths, high five!

*high fives pink oni* yeah i love english over maths anyday. I mean in past tests.

I love my english don't mind maths but if i had to choose in a life or death situation it would choose english any day

 

[ekoolish]

But it says solve by completing the square. I guess you could do it that way to check if you are right.

Yeah question isn't really that hard to waste time checking it though
<-----Random smiley.