Variable
Hello candidate,
The work done by a force is calculated as the force applied multiplied by the distance travelled by the body by that applied force in the same direction or at a specific angle.
The work done by a variable force depends as a dependence of force on the
Hello Aspirant,
Hope you are doing well!!
It is quite difficult to make integration sign, In place of integration sign I'll put |, and in place of power I'll put ** when you are solve in your notebook you can place | sign to integration sign and ** to power
Dear Student,
Dear aspirant,
Work is the energy transferred from the object via the application of force along the displacement.
Power is basically the amount of energy transferred per unit time.
Work done by a variable force is the integral of F.dx where F is the force and dx is the displacement.
Hi.
This law exhibits the short-run production functions in which one factor varies while the others are fixed.
Also, when you obtain extra output on applying an extra unit of the input, then this output is either equal to or less than the output that you obtain from the previous
Hello student,
The linear equation in one variable is an equation which can be expressed in the form such as 'ax + b = 0' (a, b and c are real numbers) where a and b are the two integers , and x has just one solution also which is
Hello aspirant,
Here is the answer to your question.
Marginal costing shows that the meaning of profit gets changed, but here profit means contribution.
Hence, Selling Price= 10 per unit-Variable Cost=6 per unit contribution which is equal to 4 per unit total contribution= 4 per unit * 100000 units= 4
The answer will be-
Assume slope of line be m
(2,3) are fixed points
Equation of line:-
y-3=m*(x-2)
y-3=mx-2m
mx-y+3-2m=0
Foot of perpendicular from origin (0,0) is-
(2-0)/m=(y-0)/(-1)= -(0+3-2m)/{0.5^(1+m^2)}
m= -x/y
Now, mx-y+3-2m=0
(-x/y)x-y+3-2(-x/y)=0
x^2+y^2-2^x-3^y=0
Centre=(1, 3/2)
Radius= 0.5^(1^2+(3/2)^2-0) = √13/2
Diameter = 2*Radius = 2* (√13/2) =
Hello!
31-(a) Transition metals show variable oxidation state because of their valence electron in two orbitals. Difference between the two orbitals is too less, hence both the energy levels can form bonds.
(b) Zinc (Zn) have higher tendency to loose electron when compared to that of Cupper(Cu)
(c) Mn shows
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