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I am sorry in a bit hurry if I get time will for sure solve it...bye...Yes, pls..?
The book states the answer is 2^5/4 x^-1 y^5/4 .. but I dunno how to reach at this answer..
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I am sorry in a bit hurry if I get time will for sure solve it...bye...Yes, pls..?
The book states the answer is 2^5/4 x^-1 y^5/4 .. but I dunno how to reach at this answer..
I understood it now Jazakillah Khair
base for natural logarithms and instead of 10, is 2.71828 function. ln on calculator gives log to this base.Aslamoalikum
need some help in exponential and logs thingy
what does basically 'e' stand for in exponential functions? i mean what it is and its significance?
JazakAllah and Thank you
i HOPE U UNDERSTOOD!!doubt in the following questions. plz someone solve it.
View attachment 14832
smzimran,
nightrider1993
THIS IS ANOTHER FOR YOU!!GT IT?doubt in the following questions. plz someone solve it.
View attachment 14832
smzimran,
nightrider1993
i HOPE U UNDERSTOOD!!
doubt in the following questions. plz someone solve it.
View attachment 14832
smzimran,
nightrider1993
Hey sorry i read your question just now. but i think parthrock's already answered it pretty well
nightrider1993 or parthrocks Please help me for the above question that I postedHey sorry i read your question just now. but i think parthrock's already answered it pretty well
Which question?
qstn 4 is implicit function so differentiate x and y both.
x^1/2 + y^1/2 = a^1/2 a is a constant means a figure that can be 2 ,3 , or else
differentiating
1/2x^(1/2-1)dx/dx + 1/2y^(1/2-1)dy/dx = O as "a" is a constant
1/2x^(-1/2) + 1/2y^(-1/2)dy/dx = O
make dy/dx the subject
1/2y^(-1/2)dy/dx = - 1/2x^(-1/2) cancel 1/2 on both sides.
y^(-1/2)dy/dx = - x^(-1/2) send y^(-1/2) on R.H.S
dy/dx = - x^(-1/2) /y^(-1/2) take inverse to make the powers positive
dy/dx = - (y/x)^1/2 Ans
have to go.. will do qstn 8 a little later
Actually, I needed help for the second part !!
x3 − x − 2Actually, I needed help for the second part !!
x3 − x − 2
(x − 1)(x2+ 1)
.
(i) Express f(x) in the form
remember both powers of numenator and denominator are equal in this part, so this is an improper fraction.
denominator when expanded in this fraction becomes
x^3 + x - x2 - 1
now divide the numenator for denominator.
as we always do in polinomial fractions.
it is shown in the attachment below.
now 1 + ( x^2 - 2x - 1 )/ (x-1)(x^2 + 1)
1 is the constant known as A here.
now rest is a proper fration.
B/(x-1) + Cx+D(X^2 + 1) = x^3 - X -2
taking lcm
B(X^2+1) + Cx+D(X-1) = X^3 - X -2
take x= 1
2B + 0 = - 2
B = -1
x= 0
B(X^2+1) + Cx+D(X-1) = X^3 - X -2
-1 -D = -1 -1 - 2
D = 3
x= -1
B(X^2+1) + Cx+D(X-1) = X^3 - X -2
-2 + (-C +3)(-2) = -1 +1 -2
-2 + 2C -6 = -2
2c -8 = -2
2c = -2+ 8
c = 3
hope these values are correct have not checked the m.s yet.
View attachment 14858
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