Tony wants to take karate lessons. the dojo is having a special monthly membership with unlimited lessons. there is an initiation fee of $20 and each month of membership costs $52.50. if tony pays $230, write and solve an equation to determine how long his membership will last.

4 months

Step-by-step explanation:

So first you subtract 20 from 230 because of the initiation fee.

230 - 20 = 210

Then you divide 210 by 52.5 to find out how many months Tony's membership will last.

210/52.5 = 4.

So, Tony's membership will last 4 months.

.there can be infinite values of m and n that can satisfy the above equation. for each value of m of m or n. there maybe infinite sets of solution for the above

step-by-step explanation:

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!