0
Want this question answered?
Be notified when an answer is posted
Add your answer:
Earn +20 pts
Why is necessary to substitute the solution back into the equation?
The equation needs an answer for it to be an equation in the 1st place. You bring the answer back to equation to show it's complete
What are the 4 steps to solving an equation?
The 1st step would be to give an example of the equation to be solved.
What are the four steps for solving a quadratic equation?
The 1st step would have been to show a particular quadratic equation in question.
How can you obtain third equation of motion from first equation?
The 1st and 3rd Equation of motion are the same, the force is zero. Thus 0 =force = Sum forces = action + reaction =0
What is the point of intersection of the lines 3x plus 2y equals 2 and x -3y equals -14?
1st equation: 3x+2y = 2 2nd equation: x-3y = -14 Multiply all terms in the 2nd equation by 3 and subtract it from the 1st equation:- So: 11y = 44 or y = 4 By substitution point of intersection is at: (-2, 4)
In first order both x and y in chemistry?
what in 1st order?? if you are asking about 1st order chemical equation then there will be only one variable with power 1.
What does the Differential on an automatic transmission do?
Differential is the 3rd member of the power train The engine is the 1st (powering the vehicle) The transmission is the 2nd member (taking that power and transmitting through various gear ratios to the differential) The differential takes that adjusted power to the drive wheels
Why is necessary to substitute the solution back into the equation?
The equation needs an answer for it to be an equation in the 1st place. You bring the answer back to equation to show it's complete
Define order of operations?
The order in which you should solve an equation to get the right answer. Use P.e.m.d.a.s. or : (1st)Parentheses, (2nd) Exponents, (3rd) Multiplication, (4th) Division, (5th) Addition, and (6th) Subtraction
What are the 4 steps to solving an equation?
The 1st step would be to give an example of the equation to be solved.
Is rate constant equal to one over TIME in the Arrhenius Equation If so or not is there an equation representing their relationship?
The unit of the rate constant in a 1st Order reaction rate equation (NOT the 'Arrhenius equation', as stated in the question) is One over Time.General form of a reaction rate equation :rate (mol.L-1.time-1) = [rate constant(Ln-1.mol1-n.time-1)]*[Concentration()]nwhere:* n is the Order of the rate equation (that is of the rate limiting step) * all units are (italicalised) between brackets It can easily be seen in this that for n=1 (1st Order) the equation is:r = k * C1and in units:mol.L-1.time-1 = (L0.mol0.time-1)*(mol.L-1)1so:(mol.L-1.time-1) = (time-1)*(mol.L-1)Only the value of the rate constant k is depending on temperature only (cf. Arrhenius equation), though temperature is NOT in its unit.
What are the four steps for solving a quadratic equation?
The 1st step would have been to show a particular quadratic equation in question.
How can you solve second order differential equation using op amp ckt?
this can be done using 3 op-amps in series.inthe 1st one connect a capacitor for feedback with a key across its end to charge and discharge it.in 2nd cnect capacitor across 2-6 and across its ends connect batery with akey in sereis.in 3rd just connect r for feedbak.conect o/p of final to 1st opamp through R.take o/p across final 6and ground
What are the advantage of state space equation representation?
There are several advantages to using the state space representation compared with other methods. Some of these are: the ability to easily handle systems with multiple inputs and outputs; the system model includes the internal state variables as well as the output variable; the model directly provides a time-domain solution, which ultimately is the thing of interest; the form of the solution is the same as for a single 1st-order differential equation; the effect of initial conditions can be easily incorporated in the solution; the matrix/vector modeling is very efficient from a computational standpoint for computer implementation
How can you obtain third equation of motion from first equation?
The 1st and 3rd Equation of motion are the same, the force is zero. Thus 0 =force = Sum forces = action + reaction =0
What is the point of intersection of the lines 3x plus 2y equals 2 and x -3y equals -14?
1st equation: 3x+2y = 2 2nd equation: x-3y = -14 Multiply all terms in the 2nd equation by 3 and subtract it from the 1st equation:- So: 11y = 44 or y = 4 By substitution point of intersection is at: (-2, 4)
What is the the point of intersection of the equations 5x plus 7y plus 29 equals 0 and 11x -3y -65 equals 0?
1st equation: 5x +7y +29 = 0 or 5x +7y = -29 2nd equation: 11x -3y -65 = 0 or 11x -3y = 65 Times all terms in the 1st equation by 11: 55x +77y = -319 Times all terms in 2nd equation by 5: 55x -15y = 325 Subtract the 2nd equation from the 1st equation: 92y = -644 or y = -7 By means of substitution point of intersection is made at: (4, -7)
