The usual rules of addition and multiplication apply to complex numbers as well as to real numbers. The true statements are ...
- x + y = y + x . . . . . . . . . . . . . . . commutative property of addition
- (x × y) × z = x × (y × z) . . . . . . . . associative property of multiplication
- (x + y) + z = x + (y + z) . . . . . . . . associative property of addition
Answer:
Ruby’s function notation V(h) describes the volume of the pool after h number of hours. Hours is the independent variable and the total volume is the dependent variable.
Step-by-step explanation:
sample response
Answer:
<h2>It must be shown that both j(k(x)) and k(j(x)) equal x</h2>
Step-by-step explanation:
Given the function j(x) = 11.6
and k(x) =
, to show that both equality functions are true, all we need to show is that both j(k(x)) and k(j(x)) equal x,
For j(k(x));
j(k(x)) = j[(ln x/11.6)]
j[(ln (x/11.6)] = 11.6e^{ln (x/11.6)}
j[(ln x/11.6)] = 11.6(x/11.6) (exponential function will cancel out the natural logarithm)
j[(ln x/11.6)] = 11.6 * x/11.6
j[(ln x/11.6)] = x
Hence j[k(x)] = x
Similarly for k[j(x)];
k[j(x)] = k[11.6e^x]
k[11.6e^x] = ln (11.6e^x/11.6)
k[11.6e^x] = ln(e^x)
exponential function will cancel out the natural logarithm leaving x
k[11.6e^x] = x
Hence k[j(x)] = x
From the calculations above, it can be seen that j[k(x)] = k[j(x)] = x, this shows that the functions j(x) = 11.6
and k(x) =
are inverse functions.