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tino4ka555 [31]
4 years ago
6

Ralph is purchasing a new gaming system for $300.00. He is trying to decide whether he wants to use cash or his credit card to p

ay. He has $600.00 available in his checking account. Which of the following is not an argument that justifies Ralph's use of his cash to purchase the gaming system? a. Using cash can help Ralph improve his credit score. b. Using cash may limit overspending or impulse buying. c. Cash doesn't come with any potential "over limit" or "late payment" fees. d. Cash will be the easiest method of payment with very few security measures involved.
Mathematics
2 answers:
ra1l [238]4 years ago
6 0

Using cash can help Ralph improve his credit score is just untrue in so many ways.


The answer is choice A because the statement does not justify Ralph's use of cash.

timurjin [86]4 years ago
6 0

Ralph is purchasing a new gaming system for $300.00. He is trying to decide whether he wants to use cash or his credit card to pay. He has $600.00 available in his checking account.

The following statement is not an argument that justifies Ralph's use of his cash to purchase the gaming system-

Using cash can help Ralph improve his credit score.

An argument can be defined as a reason or set of reasons given with the aim of persuading others that an action or idea is right or wrong.

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We can try reduction order and look for a solution y_2(x)=y_1(x)v(x). Then

y_2=y_1v\implies{y_2}'=y_1v'+{y_1}'v\implies{y_2}''=y_1v''+2{y_1}'v+{y_1}''v

Substituting these into the ODE gives

x^2(y_1v''+2{y_1}'v+{y_1}''v)-3x(y_1v'+{y_1}'v)+5y_1v=0

x^2y_1v''+(2x^2{y_1}'-3xy_1)v'+(x^2{y_1}''-3x{y_1}'+5y_1)v=0

x^4\cos(\ln x)v''+x^3(\cos(\ln x)-2\sin(\ln x))v'=0

which leaves us with an ODE linear in w(x)=v'(x):

x^4\cos(\ln x)w'+x^3(\cos(\ln x)-2\sin(\ln x))w=0

This ODE is separable; divide both sides by the coefficient of w'(x) and separate the variables to get

w'+\dfrac{\cos(\ln x)-2\sin(\ln x)}{x\cos(\ln x)}w=0

\dfrac{w'}w=\dfrac{2\sin(\ln x)-\cos(\ln x)}{x\cos(\ln x)}

\dfrac{\mathrm dw}w=\dfrac{2\sin(\ln x)-\cos(\ln x)}{x\cos(\ln x)}\,\mathrm dx

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\ln|w|=\displaystyle\int\frac{2\sin u-\cos u}{\cos u}\,\mathrm du=\int(2\tan u-1)\,\mathrm du

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w=e^{-2\ln(\cos u)-u+C}

w=Ce^{-u}\sec^2u

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