Solving Tables of Values with Equations
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
- \( q = 1 \)
- \( r = 5 \)
- \( s = 25 \)
