Carlos bought a $235 water heater with his credit card. he used the water heater for five years before replacing it. he paid off the water heater after two years, making monthly payments. the water heater cost him an average of $1.56 per week in electricity, and $0.78 per week in water. if carlos’s credit card has an apr of 14.15%, compounded monthly, and he made no other purchases with it, what percentage of the lifetime cost of the water heater was interest? (round all dollar values to the nearest cent.) a. 14.15% b. 13.35% c. 4.12% d. 30.83%

Solution:

Cost of water heater= $ 235

Carlos Credit Card APR=  14.15%, compounded monthly

Cost of water heater in terms of electricity used and water used per week(if we consider two consecutive years as a non leap year) =

Cost of water heater(if we consider two consecutive years as a  leap year)=$244.02

It is Approx $ 9 more than actual cost.

Amount that Carlos has to pay after 2 years , if credit cards APR is 14.15%, compounded monthly=

Interest Paid = $ 305.60 - $ 235=$ 70.60

As, the life of water heater is 5 years.

Life time cost of water heater=if we consider 52 × 5 +1= 261 weeks, then total cost of water heater= 261×1.56+261×0.78

= $ 610.74

The percentage of the lifetime cost of the water heater that was interest=(Approx)

Percentage of lifetime cost of water heater that was interest=11.55%

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.