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easy
product rule
this means
-e^(-x)*sin(x)+e^(-x)*cos(x)=0
now solve the equation
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easy
Thanks bro! studyingrobot457easy
product rule
this means
-e^(-x)*sin(x)+e^(-x)*cos(x)=0
now solve the equation
Anyone there to help me I’m kinda stuck in this ques that says
A curve has equation y=2x^2-2x
Find the set of values of x for which y >9
It’s from 9709/12/nov/dec/13
We have to find integral constant C hereGuys, plz help me on thisView attachment 62931
What should i do then? I am really bad when it comes to integralWe have to find integral constant C here
Ooohh.. I dont see it before haha. Thank you! It helped me a lotIve attached the solution see.. If it doesnt help then tell me
Here you go lisa ☺️ see attachmentsGuys! Plz help me! Im stuck again on integration problem. I cant answer any of the question below.View attachment 62936


still need help with that ?Please help me on functions. I forgotten all the basics.View attachment 62938
Can anyone please help me .
9709/12/O/N/17
Each year, the value of a certain rare stamp increases by 5% of its value at the beginning of the year. A collector bought the stamp for $10 000 at the beginning of 2005. Find its value at the beginning of 2015 correct to the nearest $100.
its a sequence question so i dont think they will give marks for thisVery similar to compound interest.
F = P(1 + r/100)^n
Where
F = Final amount
P = Initial amount
r = increment
n = number of years
F = 10,000(1 + 5/100)^10
F = 10,000(1.05)^10
F = 10,000(1.6289)
[ F = $16,289 ≈ $16,300 ]
Assalamu alaikum wa rahmatullah.
I picked up math again after a long time.
Pure Mathematics 1, by Hugh Neill and Douglas Quadling, Chapter 1, Miscellaneous exercise 1, questions 4, 20 and 22.
4. A(7,2) and C(1,4) are two vertices of a square ABCD.
a. Equation of the diagonal BD.
b. Coordinates of B and D.
For part a., I know the product of two diagonals is -1, but I can't get it
Q 20 is similar, but a rhombus.
22. Two lines have equations y=m1x+c1 and y=m2x+c2, and m1m2= -1. Prove that the lines are perpendicular.
(PS for 22, I couldn't type the 1s and 2s in subscript, sorry.)
Jazakumullahu khayr
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