Two different plants grow each year at different rates, which are represented by the functions f(x) = 4x and g(x) = 5x + 2. What is the first year the f(x) height is greater than the g(x) height? A. Year 0 B.Year 1 C.Year 2 D.Year 3 Can i get some help plz

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Two different plants grow each year at different rates, which are represented by the functions f(x) = 4x and g(x) = 5x + 2. What is the first year the f(x) height is greater than the g(x) height? A. Year 0 B.Year 1 C.Year 2 D.Year 3 Can i get some help plz

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@Michele_Laino can u help me plz
you have to substitute x=0, 1,2,3 then you have to compare the values f(x) and g(x) for each year. For example, if x=0, then we have: f(0) = 4*0 =0, and g(0)= 5*0+2=2, so f(0) < g(0) now, please do the same for other values of x

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so pu 4 in for f in the problem and 5 in for g
u there
yes!
question: are f(x) and g(x) heights or rates?
rates
so, are you familiar with differential calculus?
no i am not ?
if f(x) is a rate, then the corresponding height h, is given solving this differential equation: \[\Large \frac{{dh}}{{dx}} = 4x\]
ok i am trying to understand what is going on but for my first answer i put B. am i right
I'm pondering...
okay
my reasoning is: height h1 for f(x) is: \[\Large {h_1}\left( x \right) = 2{x^2}\] whereas height h2 for g(x) is: \[\Large {h_2}\left( x \right) = \frac{{5{x^2}}}{2} + 2x\] and we have h2>h1, if x>0, whereas h2=h1 if x=0
please wait I ask to another helper
so A
@Astrophysics please help
so what i need to do
This doesn't really make sense, are the functions \[f(x)=4x~~and~~g(x) = 5x+2~~~~or~~~~f(x)=4^x~~~~and~~~g(x) = 5x+2\]
Two different plants grow each year at different rates, which are represented by the functions f(x) = 4x and g(x) = 5x + 2. What is the first year the f(x) height is greater than the g(x) height? this is the problem
What are the functions...from what i showed you
is it to the power of x or just times x...
just times
So it's \[f(x) = 4x\] and \[g(x) = 5x+2\]
i am so confused tho
Can you please take an image of your question and post it
i dont know how to do thatfrom computer
Two different plants grow each year at different rates, which are represented by the functions f(x) = 4x and g(x) = 5x + 2. What is the first year the f(x) height is greater than the g(x) height? Year 0 Year 1 Year 2 Year 3
Ok all I am asking you is, what are the functions they gave you in the problem f(x), if it's to the power of x or is just 4x...
and same for g(x)
|dw:1438873919146:dw|
Two different plants grow each year at different rates, which are represented by the functions f(x) = 4x and g(x) = 5x + 2 What is the first year the f(x) height is greater than the g(x) height? Year 0 Year 1 Year 2 Year 3
cmon dude lol
lol tbh i dont know
Im asking you about the problem
not how to do it
lol o....
Two different plants grow each year at different rates, which are represented by the functions `f(x) = 4x and g(x) = 5x + 2` What is the first year the f(x) height is greater than the g(x) height?
Are those to the power of x or just being multiplied by x
being multiplied
So it's exactly \[f(x) = 4x~~~~~g(x) = 5x+2\] like that right?
yes that right
Ok well...then you can make a graph check this out |dw:1438874246367:dw| blue is g(x) and green is f(x)
what am i looking for in the picture
This doesn't make sense
This shows g(x) > f(x) always
ok so make a graph
Can you please take a screenshot of your problem
how can u screen shot from computer
did u get it
not yet
Ok I got it and it was to the power of x, I KNEW IT!! haha|dw:1438875745626:dw|
lol okay
Ok so now we can put in values as Michele did above, to see so let x = 0 \[f(0) = 4^0 = 1\] \[g(0) = 5(0)+2=2\] So its not year 0
its 2
C.
\[f(1) = 4^1=4\] \[g(1) = 5(1)+2=7\] year 1 \[f(2) = 4^2 = 16~~~~g(2) = 5(2)+2 = 12\]
Yup year 2 lol
Bleh, that took so long xD, I knew something was fishy
lol okay thx for your time
Which of the following exponential functions goes through the points (1, 6) and (2, 12)? f(x) = 3(2)x f(x) = 2(3)x f(x) = 3(2)−x f(x) = 2(3)−x
help me with this @Astrophysics
Lol no problem
Please open a new question
lol okay

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