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question:For a matrix M with rank n-1, why is it true that at least one of its minors of size n-1 is nonzero?
answer:By definition, the rank of M is the size of the largest nonzero minor. Since the rank is n-1, there exists a minor of size n-1 that is nonzero.
question:Solve the differential equation left(frac{dy}{dt} right)^{2} = frac{A^2B^2}{C^2}-y^2 (1-3y) ,, , where A,B,C are constants, using the method of separation of variables.
answer:int dt = int frac{dy}{sqrt{D^2-y^2 (1-3y)}} quad;quad D^2=frac{A^2B^2}{C^2} Using the same method as before, we find the roots y_1, y_2, y_3 of the cubic equation D^2-y^2(1-3y)=0. Then, we have: int dt = int frac{dy}{sqrt{3(y-y_1)(y-y_2)(y-y_3) }} t = -sqrt3::Fleft(Y:|: nright)+text{constant, } where F(Y:|:n) is the elliptic integral of the first kind, with begin{cases} Y=sin^{-1}left(sqrt{frac{y_2-y_1}{y-y_1}} right) n=frac{y_1-y_3}{y_1-y_2} end{cases} The inverse function is the Jacobi amplitude function, and the solution is: y(t)=y_1+frac{y_2-y_1}{sin^2left(text{am}left(-frac{t}{sqrt3}+c:bigg|: frac{y_1-y_3}{y_1-y_2}right)right)}
question:How much energy in kilojoules is required to heat 50.0 g of liquid water at a temperature of 25°C to gaseous water at a temperature of 150°C?
answer:To calculate the energy required, we need to consider the following steps: 1. Heating the water from 25°C to 100°C (boiling point) 2. Converting the water from liquid to gas (vaporization) 3. Heating the steam from 100°C to 150°C The energy required for each step can be calculated as follows: 1. Energy to heat the water: Q1 = m * C * ΔT where m is the mass of water (50.0 g), C is the specific heat of water (4.184 J/g°C), and ΔT is the change in temperature (100°C - 25°C = 75°C). Q1 = 50.0 g * 4.184 J/g°C * 75°C Q1 = 15,690 J 2. Energy to vaporize the water: Q2 = m * Lvap where Lvap is the latent heat of vaporization of water (2,260 J/g). Q2 = 50.0 g * 2,260 J/g Q2 = 113,000 J 3. Energy to heat the steam: Q3 = m * C * ΔT where C is now the specific heat of steam (2.01 J/g°C), and ΔT is the change in temperature (150°C - 100°C = 50°C). Q3 = 50.0 g * 2.01 J/g°C * 50°C Q3 = 5,025 J Therefore, the total energy required is: Qtotal = Q1 + Q2 + Q3 Qtotal = 15,690 J + 113,000 J + 5,025 J Qtotal = 133,715 J Converting to kilojoules: Qtotal = 133,715 J * (1 kJ / 1000 J) Qtotal = 133.715 kJ Hence, the energy required to heat 50.0 g of liquid water at 25°C to gaseous water at 150°C is 133.715 kJ.
question:The bank statement for Williams Company indicates a balance of 1,250 on August 31. The cash balance per books had a balance of 720 on this date. The following additional information is available: * A deposit of 810, representing cash receipts of August 31, was not included on the bank statement. * Checks written in August that have not yet cleared the bank totaled 440. * The bank charged a 25 service fee for the month of August. * The bank collected a 1,500 note receivable plus 75 interest for Williams Company. * A customer's check for 280 was returned by the bank as NSF. Prepare a bank reconciliation and any necessary adjusting entries for Williams Company as of August 31.
answer:Bank Reconciliation: | Amount | Explanation | |---|---| | 1,250 | Balance per bank, unadjusted | | 810 | Deposit in transit | | (440) | Outstanding checks | | 1,620 | Balance per bank, adjusted | Book Reconciliation: | Amount | Explanation | |---|---| | 720 | Balance per books, unadjusted | | 1,575 | Note and interest collected | | (25) | Bank service fees | | (280) | NSF check | | 1,620 | Balance per books, adjusted | Adjusting Entries: | Account | Debit | Credit | Explanation | |---|---|---|---| | Cash | 855 | | Adjust cash to correct amount (1,620 - 720) | | Bank Service Fees | 25 | | Expense bank service fees | | Accounts Receivable | 280 | | Restore receivable related to NSF check | | Note Receivable | 1,500 | | Remove note receivable collected by the bank | | Interest Revenue | 75 | | Record interest revenue earned on note |