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CaHeK987 [17]
3 years ago
12

NEED HELP ASAP!! PLEASE SHOW AND EXPLAIN YOUR ANSWER!

Mathematics
1 answer:
Evgen [1.6K]3 years ago
6 0

A. The decrease in Car 1 value is constant at 6000 from year to year. Its value can be described by a linear function.


The decrease in Car 2 value is a constant percentage of its current value from year to year. Its value can be described by an exponential function.


B. The slope of the Car 1 function is -6000, and the value is 32000 in year 1, so the value function for Car 1 can be written as

... f(x) = 32000 -6000(x-1)

... f(x) = 38000 -6000x


The value multiplier for Car 2 from year to year is 27455/32300 = 0.85, so the value function for Car 2 can be written as

... f(x) = 32,300·0.85^(x-1)

... f(x) = 38000·0.85^x


C. There is a significant difference in value after 5 years. Car 2 is worth more than double the value of Car 1 in year 5. That year, their values are ...

... car 1: $8000

... car 2: $16861

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Elodia [21]

The reduction from of the equation is \rm  -2y^2+11y-10.

<h2>Given that</h2>

Reduce the equation; \rm  7y +2y ^2 - 7 by \rm 3 - 4y.

<h3>According to the question</h3>

To reduce the given equation follow all the steps given below.

Reduce the equation; \rm  7y +2y ^2 - 7 by \rm 3 - 4y.

To reduce the equation means we need to subtract one equation from another.

Then,

The reduction from the equation is,

\rm =7y-2y^2-7-(3-4y)\\&#10;\\&#10;= 7y-2y^2-7-3+4y\\&#10;\\&#10;= 11y-2y^2-10\\&#10;\\&#10;= -2y^2+11y-10

Hence, the reduction from of the equation is \rm  -2y^2+11y-10.

To know more about Subtraction click the link is given below.

brainly.com/question/26182329

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