Mathematics: Post your doubts here!

[s1294]

change 2cos2(theta -1 = o by substituting y= cos2(theta
i.e
2y - 1 = o Y is substituted for cos2(theta now it will be y = 1/2

U can see frm graph that at how many positions is this line of y =1/2 cutting the graph (at four)

ohh okay so u substituted it.. cool.. so easy n i couldn't even see it.....

so what is the difference when the question says ' write down the number o roots' and ' deduce the number of roots' ?

when the interval changes to 10pi<= (theta)<=20pi what do i do? do i draw another graph?

 

[leadingguy]

ohh okay so u substituted it.. cool.. so easy n i couldn't even see it.....

so what is the difference when the question says ' write down the number o roots' and ' deduce the number of roots' ?

when the interval changes to 10pi<= (theta)<=20pi what do i do? do i draw another graph?

we found out 4 roots from opi to 2pi
now he is asking the no. of roots form 10pi to 20pi

look frm graph at the regular intervals of 2pi, we are having 4 roots.
how many intervals we will have from 10 to 20pi?? exactly 5 intervals
means 5 * 4 = 20 roots.

 

[s1294]

Please help here!!


from here you are told theta is pi/5 and r is 10

i have to find x and y length so that i can get the area of the triangle
i knw SOH CAH TOA has to be used so

for x=
10sin(pi/5)= 5.87785...... so is x= 5.9

then for y=
10cos(pi/5)= 8.09016.... so is y= 8.1?

n the area= 0.5*8.1*5.9= 23.895 = 23.9 cm^2 ??
please check n let me know

 

[s1294]

we found out 4 roots from opi to 2pi
now he is asking the no. of roots form 10pi to 20pi

look frm graph at the regular intervals of 2pi, we are having 4 roots.
how many intervals we will have from 10 to 20pi?? exactly 5 intervals
means 5 * 4 = 20 roots.

how do you know it will be exactly 5?? and why do u multiply by 4?

 

[umarashraf]

send me the complete q... i think i can help you mate...

 

[s1294]

send me the complete q... i think i can help you mate...

its written in this doc

 

[umarashraf]

its written in this doc

is it a right angle triangle...??? confirm

 

[s1294]

is it a right angle triangle...??? confirm

yes it is- the top corner on the left hand side is the right angle

 

[umarashraf]

do you have the answer as well...??? its 23.776 up to my calculations....

 

[umarashraf]

yes it is- the top corner on the left hand side is the right angle

do you have the answer as well...??? its 23.776 up to my calculations....

 

[s1294]

do you have the answer as well...??? its 23.776 up to my calculations....

nope i dnt have the answer- pls explain how u got the answer

 

[s1294]

nope i dnt have the answer- pls explain how u got the answer

hey i need to go, i will check ur reply tom, goodnight, n thank very much for ur help.

 

[umarashraf]

hey i need to go, i will check ur reply tom, goodnight, n thank very much for ur help.

mate its a right angle triangle with angles of 90, 36, 54. (hope you know how to convert angle from radians to degree) to find area you just need to know any two sides and a angle between them... one length is given as 10cm ... find the length of x or y...(up to know)... then put in the area formula... that is 0.5(a)(b)sin0... where a and b are any lengths a given triangle and sin0 is the angle between the two a and b....

hope it will work... if still unclear then let me know.....

 

[s1294]

mate its a right angle triangle with angles of 90, 36, 54. (hope you know how to convert angle from radians to degree) to find area you just need to know any two sides and a angle between them... one length is given as 10cm ... find the length of x or y...(up to know)... then put in the area formula... that is 0.5(a)(b)sin0... where a and b are any lengths a given triangle and sin0 is the angle between the two a and b....

hope it will work... if still unclear then let me know.....

yeah that formula is fine but to find the other side is the problem- i need to find x or y- hw do i do that?

 

[s1294]

yeah that formula is fine but to find the other side is the problem- i need to find x or y- hw do i do that?

i have another binomial question, i can do the first bit but the second- i am not sure if i am right please check n tell me if m right or wrong

I) find the first 3 terms in the expansion of (2-y)^5 in ascending powers of y

i did the following

5C0 * (2)^5 * (-y)^0 + 5C1 * (2)^4 * (-y)^1 + 5C2 * (2)^3 * (-y)^2

= 32 - 80y + 80y^2

II) use the result of part (I) to find the coefficient of x^2 in the expansion of (2 - (2x-x^2))^5

i have the following

5C2 * (2)^3 (- (2x-x^2))^2 = 80 * (-2x +x^2)^2 = 80 * (4x^2 - 2x^3 - 2x^3 +x^4) = 80 * (4x^2 - 4x^3 + x^4)

= 320x^2 - 320x^3 + 80x^4
coefficient of x^2= 320
5C1* (2)^4 (- (2x-x^2))^1 = 80 * (-2x - x^2) = -160x + 80x^2
coefficient of x^2= 80

so the total is 320 +80= 400 which is the final answer
is this correct?

 

[umarashraf]

i have another binomial question, i can do the first bit but the second- i am not sure if i am right please check n tell me if m right or wrong

I) find the first 3 terms in the expansion of (2-y)^5 in ascending powers of y

i did the following

5C0 * (2)^5 * (-y)^0 + 5C1 * (2)^4 * (-y)^1 + 5C2 * (2)^3 * (-y)^2

= 32 - 80y + 80y^2

II) use the result of part (I) to find the coefficient of x^2 in the expansion of (2 - (2x-x^2))^5

i have the following

5C2 * (2)^3 (- (2x-x^2))^2 = 80 * (-2x +x^2)^2 = 80 * (4x^2 - 2x^3 - 2x^3 +x^4) = 80 * (4x^2 - 4x^3 + x^4)

= 320x^2 - 320x^3 + 80x^4
coefficient of x^2= 320
5C1* (2)^4 (- (2x-x^2))^1 = 80 * (-2x - x^2) = -160x + 80x^2
coefficient of x^2= 80

so the total is 320 +80= 400 which is the final answer
is this correct?

yes mate.. 100% correct... by the way you have opted a long method of solving the second part....

result of first part is 32-80y+80y^2... if you compare second question with the first, instead of 2-y it is given 2-(2x-x^2)... y is in fact replaced by (2x-x^2)
s just replace y with (2x-x^2) from the answer of first part.... that is... 32-80(2x-x^2)+80(2x-x^2)... however, your answer is definitely correct...

 

[s1294]

th

yes mate.. 100% correct... by the way you have opted a long method of solving the second part....

result of first part is 32-80y+80y^2... if you compare second question with the first, instead of 2-y it is given 2-(2x-x^2)... y is in fact replaced by (2x-x^2)
s just replace y with (2x-x^2) from the answer of first part.... that is... 32-80(2x-x^2)+80(2x-x^2)... however, your answer is definitely correct...

thank you for showing me the shorter way- saves time in exams )

 

[umarashraf]

th

thank you for showing me the shorter way- saves time in exams )

most welcome

 

[s1294]

hi can anyone help me with this question

a curve is such that dy/dx= 5 - 8/x^2. the line 3y+x=17 is the normal to the curve at point P on the curve given that the x co-ordinate of P is positive , find

i) the coordinate of P
ii) the equation of the curve

to get to P i know that the normal line and the tangent which is dy/dx meet only once
so 5-8/x^2 = 17-x / 3
15- 24/x^2=17-x
2-x +24/x^2

but i know this sint right because after solving i get x^3 so i thought i have the integrate the dy/dx and then make it equal to the line

i dnt really know- can someone please help me out????

 

[s1294]

hi can anyone help me with this question

a curve is such that dy/dx= 5 - 8/x^2. the line 3y+x=17 is the normal to the curve at point P on the curve given that the x co-ordinate of P is positive , find

i) the coordinate of P
ii) the equation of the curve

to get to P i know that the normal line and the tangent which is dy/dx meet only once
so 5-8/x^2 = 17-x / 3
15- 24/x^2=17-x
2-x +24/x^2

but i know this sint right because after solving i get x^3 so i thought i have the integrate the dy/dx and then make it equal to the line

i dnt really know- can someone please help me out????

Please help me out!!!!