Mathematics: Post your doubts here!

[snowbrood]

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

http://sh.st/yH8x3 here u go hope that helped

 

[Thought blocker]

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

You mean to say, double differentiate?

 

[RoOkaYya G]

You mean to say, double differentiate?

yh ;p
but i was explaining to her wht ths n tht meant there

 

[Thought blocker]

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

With the help of Midha Q9 Thanks to her.
3(i)
1.10^n (5.20) = C
Use natural logarithm to form a linear equation in n and C:
=>n ln 1.10 + ln 5.20 = ln C
3.20^n (1.05) = C
=> n ln 3.20 + ln 1.05 = ln C
The two equations can now be equated:
n ln 1.10 + ln 5.20 = n ln 3.20 + ln 1.05
Rearrange the equation and you'll get n = 1.50 (correct to 3 sig. fig.)
Therefore, C = 3.20^1.50 (1.05) = 6.00 (3 s.f.)

9(i)

 

[bludlynsolja]

slms...
can someone help me with mechanics w09 pp 41 question 7, the last part.
i somwhoq dont get the part of the constant so if someone were to help, i would really appreciate.
thx.

 

[RoOkaYya G]

slms...
can someone help me with mechanics w09 pp 41 question 7, the last part.
i somwhoq dont get the part of the constant so if someone were to help, i would really appreciate.
thx.

link plz

 

[bludlynsolja]

i am so sorry am very new to this community section in xtremepapers but someone asked me to send the link to the question that was bothering me, here we go...
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w09_qp_41.pdf
it is question number 7 the second part, please explain and especially more so about the constant when integrating.
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w09_ms_41.pdf
above is the link to the mark scheme as well just for ur reference.
any help would really be appreciated thx.

 

[RoOkaYya G]

i am so sorry am very new to this community section in xtremepapers but someone asked me to send the link to the question that was bothering me, here we go...
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_qp_41.pdf
it is question number 7 the second part, please explain and especially more so about the constant when integrating.
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_ms_41.pdf
above is the link to the mark scheme as well just for ur reference.
any help would really be appreciated thx.

differenciate v to obtain a
equate tht expression obtained to the negative answer from (i)
solve for t

 

[bludlynsolja]

thx a lot!! i did that..
i M SO sorry i meant the third part and not the second.
its just now that i realised that i had already done part 2 after follwoing ur instrctions and realising that i had done that part aleady..
thx anyway, but part 3 is the one that is bugging me.
the one of 6 mrks.

 

[Emma23]

With the help of Midha Q9 Thanks to her.
3(i)
1.10^n (5.20) = C
Use natural logarithm to form a linear equation in n and C:
=>n ln 1.10 + ln 5.20 = ln C
3.20^n (1.05) = C
=> n ln 3.20 + ln 1.05 = ln C
The two equations can now be equated:
n ln 1.10 + ln 5.20 = n ln 3.20 + ln 1.05
Rearrange the equation and you'll get n = 1.50 (correct to 3 sig. fig.)
Therefore, C = 3.20^1.50 (1.05) = 6.00 (3 s.f.)

9(i)
View attachment 47995 View attachment 47996

Thank you both so much

 

[Lily9605]

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.

Thanks I got it..how about question 2

 

[RoOkaYya G]

thx a lot!! i did that..
i M SO sorry i meant the third part and not the second.
its just now that i realised that i had already done part 2 after follwoing ur instrctions and realising that i had done that part aleady..
thx anyway, but part 3 is the one that is bugging me.
the one of 6 mrks.

u should do it in 2 parts
i)
for 0<t<10 use kinematics formula
a= value in part (i)
t=10s
u=0 (from rest)
v= replace t=10 in formula for v

use s=ut + 0.5 *a*t^2
replace the values a ,t ,,u and v to obtain s (name it s1)
ii)
for 10<t<20
integrate to v to get expression for s (displacement)
replace the values with limits 10 and 20
name this s2

for displacement add s1 and s2

 

[Thought blocker]

Thanks I got it..how about question 2

Lol! Its just a modulus question even RoOkaYya G ccan do it.

 

[RoOkaYya G]

Lol! Its just a modulus question even RoOkaYya G ccan do it.

yh -________-

Thanks I got it..how about question 2

let 3^x = y

2|y-1| = y
square on both sides
2^2 (y-1)^2 = y^2
4( y^2 - 2y + 1) = y^2

solve for y

therefore 3 ^x = each answer u got for y above
ln on both side
solve for x
x=-0.369
or x=0.621

 

[Thought blocker]

Thanks I got it..how about question 2

yh -________-

let 3^x = y

2|y-1| = y
square on both sides
2^2 (y-1)^2 = y^2
4( y^2 - 2y + 1) = y^2

y=4
therefore 3 ^x =4
ln on both side
solve for x
x=-0.369

2|3^x - 1| = 3^x
2(3^x -1) = ±3^x

2(3^x -1) = +3^x and 2(3^x -1) = -3^x
2x3^x - 2 = 3^x
2x3^x - 3^x = 2
3^x = 2
log 3^x = log 2
x log 3 = log 2
x = log2/log3
x = 0. 621

2(3^x -1) = -3^x
2x3^x - 2 = -3^x
2x3^x + 3^x = 2
3x3^x = 2
3^x = 2/3
x log 3 = log 2/3
x = log(2/3) / log 3
x = -0.369

 

[Lily9605]

yh -________-


let 3^x = y

2|y-1| = y
square on both sides
2^2 (y-1)^2 = y^2
4( y^2 - 2y + 1) = y^2

solve for y

therefore 3 ^x = each answer u got for y above
ln on both side
solve for x
x=-0.369
or x=0.621

Thank you

 

[Lily9605]

2|3^x - 1| = 3^x
2(3^x -1) = ±3^x

2(3^x -1) = +3^x and 2(3^x -1) = -3^x
2x3^x - 2 = 3^x
2x3^x - 3^x = 2
3^x = 2
log 3^x = log 2
x log 3 = log 2
x = log2/log3
x = 0. 621

2(3^x -1) = -3^x
2x3^x - 2 = -3^x
2x3^x + 3^x = 2
3x3^x = 2
3^x = 2/3
x log 3 = log 2/3
x = log(2/3) / log 3
x = -0.369

Thank you!

 

[Thought blocker]

Thank you

Thank you!

What were you not getting in this question?

 

[RoOkaYya G]

i bet she dint

What were you not getting in this question?

read urs fully
its too lengthy

 

[Lily9605]

What were you not getting in this question?

I got confused when I saw 3^x ..what can I say? I'm not good in math haha..