Asked by INeedThatA$AP,NoRocky
The general form of an exponential equation is y=ab^x+k. What is the general form of the following equation?
y−3=4(5^2x+1)
Responses:
y=20(25)x+3y
y=4(25)x+4y
y=4(125)x+3y
y=20(400)x+3
y−3=4(5^2x+1)
Responses:
y=20(25)x+3y
y=4(25)x+4y
y=4(125)x+3y
y=20(400)x+3
Answers
Answered by
INeedThatA$AP,NoRocky
Bruh...what... Thats wrong Bot. Can't rely on you for nothing smh.
Answered by
oobleck
AAAaannndd the bot gets it wrong yet again!
Of course, all the choices are also incorrect
y−3=4(5^2x+1)
y = 4(5^2x+1)+3
= 4(25^x+1)+3
= 4*25^x + 7
unless maybe you meant
y−3=4(5^(2x+1))
y = 4*5^(2x+1)+3
= 4*5*25^x + 3
= 20*25^x + 3
Of course, all the choices are also incorrect
y−3=4(5^2x+1)
y = 4(5^2x+1)+3
= 4(25^x+1)+3
= 4*25^x + 7
unless maybe you meant
y−3=4(5^(2x+1))
y = 4*5^(2x+1)+3
= 4*5*25^x + 3
= 20*25^x + 3
Answered by
INeedThatA$AP,NoRocky
OH bet. Thanks man! Much appreciated!
Answered by
Bot
y
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