Mathematics: Post your doubts here!

[Thought blocker]

This question! ii) onwards.

It takes time to open image. #SlowNet
Can you post the link?

 

[lxelle]

N

It takes time to open image. #SlowNet
Can you post the link?

No link though. It's my own ques.

 

[Thought blocker]

N

No link though. It's my own ques.

What is it?

 

[lxelle]

I've found i)

 

[Thought blocker]

Hi , I've a doubt in question 10 part 3.
Why do we have to use sine of the angle and not cosine?

A normal to the plane is
<1,2,2>
and the line is in the direction
<a,2,1>.

Since the line meets the plane at the angle arctan(2), it must meet the normal at either arctan(1/2) or its supplemental angle. Since the cosine of a supplementary angle is the opposite of the cosine of the angle, we want
cos( arctan(1/2) ) = 2/sqrt(5)
or its negative.

Now, use the inner product:
<1,2,2> . <a,2,1> = a + 6
|<1,2,2>| = 3
|<a,2,1>| = sqrt(a^2 + 5)
and
cos(angle) = (+ or -) 2/sqrt(5)

Therefore
| a + 6 | = 3 * sqrt(a^2 + 5) * 2/sqrt(5).

Squaring both sides:
(a+6)^2 = 36/5 * (a^2 + 5)

Then, simplifying and factoring:
0 = 31a^2 - 60a
0 = a(31a - 60)

so
a = 0
or
a = 60/31.

 

[lxelle]

the sum of the first three terms of a geometric series is 64 and the sum of the next three terms is 27. Find the exact value of sum to infinity of the series. Thought blocker

 

[Kamihus]

the sum of the first three terms of a geometric series is 64 and the sum of the next three terms is 27. Find the exact value of sum to infinity of the series. Thought blocker

According to the sum of n terms formula: 64=a(1-r^3)/1-r
By adding the first three and next three we get the second equation: 91=a(1-r^6)/1-r
The 'a' in the numerator and 1-r in the denominator is common in the two equations so we cancel that. We get: (64/1-r^3)=(91/1-r^6) Form an equation: 64r^6-91r^3+27=0
Solve to get r=3/4 or 1.
Put it in any of the two equations to get a=27.68
According to the sum to infinity formula: 27.68/(1-0.75)=110.7

 

[lxelle]

According to the sum of n terms formula: 64=a(1-r^3)/1-r
By adding the first three and next three we get the second equation: 91=a(1-r^6)/1-r
The 'a' in the numerator and 1-r in the denominator is common in the two equations so we cancel that. We get: (64/1-r^3)=(91/1-r^6) Form an equation: 64r^6-91r^3+27=0
Solve to get r=3/4 or 1.
Put it in any of the two equations to get a=27.68
According to the sum to infinity formula: 27.68/(1-0.75)=110.7

Khob khun ka!

 

[lxelle]

According to the sum of n terms formula: 64=a(1-r^3)/1-r
By adding the first three and next three we get the second equation: 91=a(1-r^6)/1-r
The 'a' in the numerator and 1-r in the denominator is common in the two equations so we cancel that. We get: (64/1-r^3)=(91/1-r^6) Form an equation: 64r^6-91r^3+27=0
Solve to get r=3/4 or 1.
Put it in any of the two equations to get a=27.68
According to the sum to infinity formula: 27.68/(1-0.75)=110.7

QUESTION. how do you find the r from the equation? You can't use the calculator

 

[Kamihus]

QUESTION. how do you find the r from the equation? You can't use the calculator

Put it in the quadratic formula to get r^3=27/64. Then cube root to get 3/4.

 

[lxelle]

Put it in the quadratic formula to get r^3=27/64. Then cube root to get 3/4.

it is given that f(x) = 1/x^3 - x^3, for x>0. show that f is a decreasing function.

 

[Kamihus]

it is given that f(x) = 1/x^3 - x^3, for x>0. show that f is a decreasing function.

Differentiate this and then differentiate that too. Then I'll tell you what to do.

 

[Haya Ahmed]

How to solve this question anyways the answers are .. a) a=2 b=1 c=2 b) pi/12 and 5pi/12

 

[Emma23]

Can someone explain question 3 part 1 and question 9 part 1?

 

[RoOkaYya G]

How to solve this question anyways the answers are .. a) a=2 b=1 c=2 b) pi/12 and 5pi/12

a)
a=2 (it show by how many units the graph moved along y axis)
b=1 (pi/period)
c=y intercept

b)
2.5=2+sin2x
sin2x=0.5
2x=sin inverse 0.5
2x =0.52 , 3.065 using CAST quadrant. its in 1st n 2nd quadrant.
x=0.26,1.83
x=pi/12 , 7pi/12 I GOT 7PI/12

 

[lxelle]

Differentiate this and then differentiate that too. Then I'll tell you what to do.

This, that what??? Anyways, I got the answer.

 

[Andhikasm]

I am in hurry, cant edit post with symbols, hope you understand. :¬

Let the random variable X be normally distributed where with u(meu) n s(sigma) as mean n standard deviation respectively....
Nw we need to find
P((u+s)<X<(u-s))

Now standarising
we have
p{((u-s-u)/s)<Z<((u-s-u)/s)}
so, P(-1<Z<1)
so phi(1)-phi(-1)
=phi 1 -[1-phi(-1)]
={2phi(1)}-1
=(2*0.8413)-1
= .6826

therefore for 1 observation the p is .6862
so, for 800 it is .6826*800
=546 (Ans)

Thanks a lot buddy, I really appreciate the help

 

[RoOkaYya G]

This, that what??? Anyways, I got the answer.

she means differenciate tht equation n then u differenciate the an answer u get again

 

[lxelle]

she means differenciate tht equation n then u differenciate the an answer u get again

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w09_qp_12.pdf 10 c please.

 

[RoOkaYya G]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_qp_12.pdf 10 c please.

find gradient at A or B for both line and curve
suppose u choose A
find gradient at A for line..lets say its k
find gradient at A for curve...lets say its l

for line : tan theta = k
theta = a

for curve : tan theta = l
theta = b

angle between tangent and line = a- b = answer