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
4 answers
y
Bruh...what... Thats wrong Bot. Can't rely on you for nothing smh.
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
OH bet. Thanks man! Much appreciated!
1 answer