Question

Complete the tables of values.

Answer 1

Answer:

for $f(x) =4^{-x}$ the value $a=1,\ b=\frac{1}{16},\ c=\frac{1}{256}$

for  $f(x) =(\frac{2}{3})^{x}$ the value $a=1,\ b=\frac{4}{9},\ c=\frac{16}{81}$

Step-by-step explanation:

Consider

1 ) $f(x) =4^{-x}$

We have to find the values at certain given points.

at x = 0

$f(0) =4^{-0}=4^0=1$

at x = 2

$f(2) =4^{-2}=\frac{1}{4^2}=\frac{1}{16}$

at x= 4

$f(4) =4^{-4}=\frac{1}{4^4}=\frac{1}{256}$

Thus, $a=1,\ b=\frac{1}{16},\ c=\frac{1}{256}$

Consider

2)  $f(x) =(\frac{2}{3})^{x}$

We have to find the values at certain given points.

at x = 0

$f(x) =(\frac{2}{3})^{0}=1$

at x = 2

$f(x) =(\frac{2}{3})^{2}=(\frac{4}{9})$

at x = 4

$f(x) =(\frac{2}{3})^{4}=\frac{16}{81}$

Thus, $a=1,\ b=\frac{4}{9},\ c=\frac{16}{81}$

Answer 2

Answer with Step-by-step explanation:

We are given two tables and we are to complete the missing values.

For table one, we are to substitute the given values of x in the term $4^{-x}$:

when x = 0, $4^{-0} = 1$

when x = 2, $4^{-2} = \frac{1}{16}$

when x = 4, $4^{-4} = \frac{1}{256}$

For table two, we are to substitute the given values of x in the term $(\frac{2}{3} )^x$:

when x = 0, $(\frac{2}{3} )^0 = 1$

when x = 2, $(\frac{2}{3} )^2 = \frac{4}{9}$

when x = 4, $(\frac{2}{3} )^4 = \frac{16}{81}$