Here's the question you clicked on:
stardust3829
p(a)=.05a t(p)=3p t(a)=?
please be more detailed while posting questions.
the variables are just variables, they don't mean anything...
\[p(a)=0.05a\]\[t(p)=3p\] Then \[t(a)=(3)(0.05a)\]\[t(a)=0.15a\]
How do you know how to combine the functions?
t(p)=3p t(1) = 3*1 = 3 t(2) = 3*2 = 6 t(a) = 3*a = 3a
To be More Specific: f(x) = 2x + 3 f(1) = 2+3 f(a) = 2a + 3
You haven't given us any info to say not to.. So my assumption is that p(a)=0.05a is the function way of writing p=0.05a. Therefore you can substitute that into t(p) to get \[t(a)=3(0.05a)=0.15a\]
@stardust3829 DId u get my point
but @Yahoo! also has a point. Because of the lack of information this can be interpreted in different ways.
not really, a little bit, but not much :) sorry
how does knowing what the functions are saying, help to know how to combine them?
You've given us two functions and you want us to find a third function which returns the same variable as one of the functions but takes a different input. <<<--- confusing sentence. y=2x is the same as y(x)=2x. x being your input and y your output. now if you had say h=5y. that's the same as saying h(y)=5y... with y as your input and h your output. Now you want to find h(x). which has an input of x but an output of y. so from that you know you nee to combine the equations. you get h(y)=h(2x) and since your h=5y.. y the input. you get h(2x)=5(2x). but you want it as h(x). so you simplify the right side 5(2x)=10x.. and now your h=10x, which is the same as h(x)=10x