Aku bisa mengisi perutmu kala lapar tapi aku juga membantu meredakan hawa panas

y - value x - year
[24000=a\cdot3+b\ 19000=a\cdot5+b\\ 24000=3a+b\ -19000=-5a-b\ ---------\ 5000=-2a\ a=-2500\\ b+3\cdot(-2500)=24000\ b-7500=24000\ b=31500\\ \boxed{y=-2500x+31500}
]
a) f ( x ) = − 2500 x + 31500 f ( x + 1 ) = − 2500 ( x + 1 ) + 31500 f ( x + 1 ) = − 2500 x − 2500 + 31500 f ( x + 1 ) = − 2500 x + 29000 f ( x ) − f ( x + 1 ) = − 2500 x + 31500 − ( − 2500 x + 29000 ) f ( x ) − f ( x + 1 ) = − 2500 x + 31500 + 2500 x − 29000 f ( x ) − f ( x + 1 ) = 2500 ​
b) f ( 0 ) = − 2500 ⋅ 0 + 31500 = 31500 ​
c) f ( 10 ) = − 2500 ⋅ 10 + 31500 f ( 10 ) = − 25000 + 31500 f ( 10 ) = 6500 ​

Answered by konrad509 | 2024-06-10

The car depreciates by $2,500 each year, and its purchase price was 31 , 500. T h e l in e a re q u a t i o n re p rese n t in g t h ec a r ′ s v a l u eo v er t im e i s y = -2500x + 31500$. If the car continues to depreciate at this rate, its value at year 10 will be $6,500.
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Answered by konrad509 | 2024-10-01

Jawaban:bisa es bisa juga sopPenjelasan:maaf klo salah ya kak

Answered by anggunaulyasalsabila | 2025-07-02