Establishing codomain, r2 to r3, constructing inverse

Related BrainMass Solutions

• matrix equation

  R1 / 24 -> R1 Now multiply row 1 by -40 and add that to row 2. -40R1 + R2 -> R2 Multiply row 1 by -150 and add to row 3. -150R1 + R3 -> R3 Now multiply row 2 by -3/10. -3/10R2 -> R2 Now multiply row 2 by -5/6 and add it to row 1.

• Gradient of potential energy equals minus the force

  We have: -dU/dr3 = -d/dr3 U13(r1-r3) - d/dr3 U23(r2-r3) - d/dr3 U34(r3-r4) - d/dr3 U3(r3) = We have seen in problem 1) that the above derivatives are the forces exerted on particle 3 in case of the inverse square law force.

• Resistors - Finding the Current

  Ohm's law states that the current vis a resistor is proportional to the potential **across the resistor** and inverse proportionl to the resistance: I=V/R In this case, the potential difference across R2 is NOT 15V, and the potential difference

• Discrete Mathematics : 11 Definitions - Boolean variable, Bit operation, Tautology, Contingency, Axiom, Paradox, Venn Diagram, Infinite Set, Domain of f, Codomain of f and Inverse of f

  Codomain of f: A set within which the values of a function lie. 11. Inverse of f: This may be explained several ways. The inverse of a function f(x) is typically denoted f-1(x) such that f(f-1(x))= f-1((f(x))=x.

• Difference between mapping R3 to R2 and the reverse

  The difference between R3->R2 and R2->R3 is that they have different domain and range. For the mapping R3->R2, each input is in R3 and has the form (x1,x2,x3). The output is in R2 and has the form (y1,y2).

• Lines, graphing, and break-even point

  1 0 0 1 -3 0 0 1 R3-R2 to R3 1 0 5 -5 -1 0 0 1 -4 4 1 0 0 0 1 -4 -1 1 R1-5R3 to R1 1 0 0 15 4 -5 0 1 -4 4 1 0 0 0 1 -4 -1 1 R2+4R3 to R2

• Bootstrapping and Interest Rate Example Problem

  To bootstrap the two-year rate, you can set up the following equation for the price (P) of the coupon bond: P=C_1/(1+r_1 )+(C_2+Par value)/(1+r_2 )^2 Because you can observe all of the variables except r2, the spot rate for two years, you can solve

• Circuit Nodal Analysis

  Current at R1 is equal to the sum of currents at R2 and R3, by KCL. Assuming that R1 is 4A while R2 is 2A, we have: 2.

• Superposition Theorem - Independent Sources

  divider concept, V = R4 * [ Zleft / (Zleft + R4) ] = Vv--- (2) Now, Z = (R1*RL)/(R1+RL) + (R3*R4)/(R3+R4) + R2 Zleft = [ (R1*RL)/(R1+RL) +R2 ] || R3 = {R3*[ (R1*RL)/(R1+RL) +R2 ] } / [R3+ (R1*RL)/(R1+RL) +R2] Hence by superposition