Let,
f(x) = -2x+34
g(x) = (-x/3) - 10
h(x) = -|3x|
k(x) = (x-2)^2
This is a trial and error type of problem (aka "guess and check"). There are 24 combinations to try out for each problem, so it might take a while. It turns out that
g(h(k(f(15)))) = -6
f(k(g(h(8)))) = 2
So the order for part A should be: f, k, h, g
The order for part B should be: h, g, k f
note how I'm working from the right and moving left (working inside and moving out).
Here's proof of both claims
-----------------------------------------
Proof of Claim 1:
f(x) = -2x+34
f(15) = -2(15)+34
f(15) = 4
-----------------
k(x) = (x-2)^2
k(f(15)) = (f(15)-2)^2
k(f(15)) = (4-2)^2
k(f(15)) = 4
-----------------
h(x) = -|3x|
h(k(f(15))) = -|3*k(f(15))|
h(k(f(15))) = -|3*4|
h(k(f(15))) = -12
-----------------
g(x) = (-x/3) - 10
g(h(k(f(15))) ) = (-h(k(f(15))) /3) - 10
g(h(k(f(15))) ) = (-(-12) /3) - 10
g(h(k(f(15))) ) = -6
-----------------------------------------
Proof of Claim 2:
h(x) = -|3x|
h(8) = -|3*8|
h(8) = -24
---------------
g(x) = (-x/3) - 10
g(h(8)) = (-h(8)/3) - 10
g(h(8)) = (-(-24)/3) - 10
g(h(8)) = -2
---------------
k(x) = (x-2)^2
k(g(h(8))) = (g(h(8))-2)^2
k(g(h(8))) = (-2-2)^2
k(g(h(8))) = 16
---------------
f(x) = -2x+34
f(k(g(h(8))) ) = -2*(k(g(h(8))) )+34
f(k(g(h(8))) ) = -2*(16)+34
f(k(g(h(8))) ) = 2
What is y={-\dfrac{1}{3}}x-9y=− 3 1 x−9y, equals, minus, start fraction, 1, divided by, 3, end fraction, x, minus, 9 written i
Kobotan [32]
Answer:
Step-by-step explanation:
We are given that
We have to find the standard form of given equation
By using multiplication property of equality
We know that
Standard form of equation
Therefore, the standard form of given equation is given by
