hey coke baby can u answer your question because to be honest with u if u start i wud maybe get an idea how to do it. I think that one is a challenging one
everybody!!! we get a question for mocks and i was abit confused. a little help here!!
Oil moves thru a pipeline such dat the distance 's'it moves and the time 't' are related by s^3-t^2 = 7t. find the velocity of the oil for s = 4.01m and t = 5.25s
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12 comments:
When the height of liquid in a tub is x meters, the volume of liquid is Vm3, where
V = 0.05[(3x + 2)^3 - 8]
a) find an expression for dv/dx
If the liquid enters at a constant rate of 0.081m^3s^-1
b)Find the rate at which the height of the liquid is increasing when v = 0.95
since nobody want to post up question we go start easy
2x^4 + 34x^3 + 6
differentiate this blogbakes
y = 2x^4 + 34x^3 +6
dy/dx = 8x^3 + 102x^2
hey coke baby can u answer your question because to be honest with u if u start i wud maybe get an idea how to do it.
I think that one is a challenging one
DOH TAKE IT D WRONG WAY BUT IT WASNT MEANT TO BAD
moose you is the real bake
Le me give you one
4x/5^2 + 45/6^5 +54
well, for the first question posted: dv/dx would be
:dv/dx = 0.05[9(3x + 2)^2]
right?
Prove that the function of y = x^3 , where x=a is 3a^2
What is the gradient of the curve y=√x+2 at x=16
Differentiate 7√x^3
everybody!!!
we get a question for mocks and i was abit confused. a little help here!!
Oil moves thru a pipeline such dat the distance 's'it moves and the time 't' are related by
s^3-t^2 = 7t. find the velocity of the oil for s = 4.01m and t = 5.25s
ok cruiser
4x/5^2 + 45/6^5 + 54
dy/dx = 4/25
voila!!!
cruiser you sure this is the question u want me to solve it looking kinda funny check it over na
ok tweety next one
3/2x^1/2 + 23x^2/3 + x^-3/4
take a crack at this na
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