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http://sh.st/yH8x3 here u go hope that helped
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http://sh.st/yH8x3 here u go hope that helped
You mean to say, double differentiate?she means differenciate tht equation n then u differenciate the an answer u get again
yh ;pYou mean to say, double differentiate?
With the help of Midha Q9 Thanks to her.Can someone explain question 3 part 1 and question 9 part 1?
link plzslms...
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.
differenciate v to obtain ai 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.
Thank you both so muchWith 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
Thanks I got it..how about question 2A 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.
u should do it in 2 partsthx 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.
Lol! Its just a modulus question even RoOkaYya G ccan do it.Thanks I got it..how about question 2
yh -________-Lol! Its just a modulus question even RoOkaYya G ccan do it.
Thanks I got it..how about question 2
Thanks I got it..how about question 2
2|3^x - 1| = 3^xyh -________-
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
Thank youyh -________-
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!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
What were you not getting in this question?Thank you!
read urs fullyWhat 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..What were you not getting in this question?
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