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Elina [12.6K]
2 years ago
7

How much is House A worth after two years? Round your answer to the nearest dollar.

Mathematics
2 answers:
Gekata [30.6K]2 years ago
4 0

Answer:

see below

Step-by-step explanation:

Take the original value times the appreciation and add the original value

Do this for each year

House A

year 1:

 new value = 125260+ 125260* .05

                    =125260 +6263

                     =131523

year 2:

 new value = 131523+131523* .05

                    131523+6576.15

                         138099.15

Nearest dollar  138099

House B

year 1:

 new value = 120160+ 120160* .06

                    =120160 +7209.60

                     =127369.6

year 2:

 new value = 127369.60+127369.6* .05

                    127369.60+7642.18

                         135011.78

Nearest dollar  135012

slega [8]2 years ago
4 0

Answer:

A: $138,099

B: $135,012

Step-by-step explanation:

A: growth rate: (100+5)/100 = 1.05

125260 × 1.05² = 138,099.15

B: growth rate: (100+6)/100 = 1.06

120160 × 1.06² = 135,011.776

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The number of pollinated flowers as a function of time in days can be represented by the function. f(x)
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The average increase in the number of flowers pollinated per day between days 4 and 10 is <u>39</u>, given that the number of pollinated flowers as a function of time in days can be represented by the function f(x) = (3)^{\frac{x}{2} }.

In the question, we are asked for the average increase in the number of flowers pollinated per day between days 4 and 10, given that the number of pollinated flowers as a function of time in days can be represented by the function f(x) = (3)^{\frac{x}{2} }.

To find the average increase in the number of flowers pollinated per day between days 4 and 10, we use the formula {f(10) - f(4)}/{10 - 4}.

First, we find the value of the function f(x) = (3)^{\frac{x}{2} }, for f(10) and f(4).

f(x) = (3)^{\frac{x}{2} }\\\Rightarrow f(10) = (3)^{\frac{10}{2} }\\\Rightarrow f(10) = 3^5 = 243

f(x) = (3)^{\frac{x}{2} }\\\Rightarrow f(4) = (3)^{\frac{4}{2} }\\\Rightarrow f(10) = 3^2 = 9

Thus, the average increase

= {f(10) - f(4)}/{10 - 4},

= (243 - 9)/(10 - 4),

= 234/6

= 39.

Thus, the average increase in the number of flowers pollinated per day between days 4 and 10 is <u>39</u>, given that the number of pollinated flowers as a function of time in days can be represented by the function f(x) = (3)^{\frac{x}{2} }.

Learn more about the average increase in a function at

brainly.com/question/7590517

#SPJ4

For complete question, refer to the attachment.

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