Question - Solving Tables of Values with Equations
Solution:
Sure, let's solve these tables of values one by one:For the first table associated with the equation $$ y = \frac{4}{x} $$:1. When $$ x = -1 $$, $$ y = \frac{4}{-1} = -4 $$. So, $$ a = -4 $$.2. When $$ x = 0 $$, the value for $$ y $$ is undefined because division by zero is not allowed. Therefore, $$ b $$ does not exist or is undefined.3. When $$ x = 1 $$, $$ y = \frac{4}{1} = 4 $$. So, $$ c = 4 $$.For the second table associated with the equation $$ d = 5^e $$:4. When $$ e = -1 $$, $$ d = 5^{-1} = \frac{1}{5} $$. So, $$ p = \frac{1}{5} $$ or 0.2.5. When $$ e = 0 $$, $$ d = 5^0 = 1 $$. So, $$ q = 1 $$.6. When $$ e = 1 $$, $$ d = 5^1 = 5 $$. So, $$ r = 5 $$.7. When $$ e = 2 $$, $$ d = 5^2 = 25 $$. So, $$ s = 25 $$.Putting all these values together:For the first table:- $$ a = -4 $$- $$ b $$ is undefined- $$ c = 4 $$For the second table:- $$ p = \frac{1}{5} $$ or 0.2- $$ q = 1 $$- $$ r = 5 $$- $$ s = 25 $$
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