Answer:
<h3>
ln (e^2 + 1) - (e+ 1)</h3>
Step-by-step explanation:
Given f(x) = ln and g(x) = e^x + 1 to get f(g(2))-g(f(e)), we need to first find the composite function f(g(x)) and g(f(x)).
For f(g(x));
f(g(x)) = f(e^x + 1)
substitute x for e^x + 1 in f(x)
f(g(x)) = ln (e^x + 1)
f(g(2)) = ln (e^2 + 1)
For g(f(x));
g(f(x)) = g(ln x)
substitute x for ln x in g(x)
g(f(x)) = e^lnx + 1
g(f(x)) = x+1
g(f(e)) = e+1
f(g(2))-g(f(e)) = ln (e^2 + 1) - (e+ 1)
An expression must contain a variable
The equation y-4=1 has many solution(s).
The answer is a since u do 4/5 put the whole number one then leave change flip then simplify
Answer:
Step-by-step explanation:
Let x = gallons consumed by car 1
Let y = gallons consumed by car 2
We set up our equations:
35x+40y = 1850 eq1
x+y = 50 eq2
Substituting eq 2 into eq 1,
35x+40(50-x) = 1850
35x+2000-40x = 1850
-5x+2000 = 1850
-5x = -150
x = 30
Substitute value of x into eq 2.
x+y = 50
30+y = 50
y = 20
Car 1 consumes 30 gallons of gas.
Car 2 consumes 20 gallons of gas.
