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not now in a few hourswooww mshallahexplain plz
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not now in a few hourswooww mshallahexplain plz
I am bad in Integration and differential equationsits so e z
sure tytnot now in a few hours
where u went from inbox ?sure tyt
Thanks man[ (1 + sinx)/cosx ] + [ cosx/(1 + sinx) ]
Take L.C.M.
[ (1 + sinx)(1 + sinx) + (cosx)(cosx) ] / cosx(1 + sinx)
Expand numerator only
[ 1 + 2sinx + sin²x + cos²x ] / cosx(1 + sinx)
Now simplify (sin²x + cos²x = 1)
(1 + 2sinx + 1) / cosx(1 + sinx)
(2 + 2sinx) / cosx(1 + sinx)
Factorize numerator
2(1 + sinx) / cosx(1 + sinx)
1 + sinx cancels
2/cosx
Find first derivative equate to zero get the x coordinate substitute the x coordinate in the equation given to get y coordinateOh god i forgot, how to find maximum and minimum point of a function.
I would love simple details like 'find the derivative and then equal it to zero..bla bla'
To know the if the graph has maximum / minimum or both you have to find the first derivative and then something
and to know the points of those maximum and minimum you have to find second derivative and then...
I forgot and i just can't find in my text book. Text book really confuses me a lot.
Thanks in Advance!
ok no problemI told you if i knew the answer ! . sorry but I failed to get the correct answer I will ask for someone's help
(2 + k)² = 1http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s09_ms_1.pdf
Why in question 2 k is greater than 2 but less than -6 explain
its solved buddyok no problem
psychiatrist Idk one question ZaqZainab solve 7b) 2012 mayjune variant 2, and I am bored with p1, I'll upload other two on 2nd may, I am busy with chemistry.
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in All GP the first term is a and second is ar and the third term is ar^2Can someone please solve this, step by step:
In a geometric progression, the second term is 9 less than the first term. The sum of the second and third terms is 30. Given that all the terms of the progression are positive, find the first term.
Thank u I didnt know abt the ar^2 thingin All GP the first term is a and second is ar and the third term is ar^2
as they have mentioned the sum of second and third term is 30
(ar)+(ar^2)=30
in this GP the first term is 'a' second is (a-9) and third is not given so lets just take ar^2 as its true for all
and as second term is 'ar' and also 'a-9'
ar=a-9
(ar)+(ar^2)=30
you have 2 equations
lets make a the subject and substitute it in the 2nd equation
a-ar=9
a(1-r)=9
a=9/(1-r)
not substituting
((9/(1-r))*r) + ((9/(1-r))r^2=30
(9r/(1-r))) + (9r^2/(1-r))=30
as denominators are the same
(9r+9r^2)/(1-r)=30
9r+9r^2=30-30r
9r^2 +9r +30r -30 =0
9r^2 + 39r -30=0
r=2/3 ,-5
if all terms are positive how can r be negative? so we won't take r=-5
a*(2/3) +a=9
a=27 i hope it helps
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