`x=36z`
`y=36z^(2)+5`
If `z gt 0` in the equations above, what is y in terms of x?

Options

A. `y=(1)/(36)x^(2)+4`
B. `y=(1)/(36)x^(2)+5`
C. `y=(1)/(36)x^(2)+36`
D. `y=(1)/(6)x^(2)+4`

Answer

To find \( y \) in terms of \( x \) given the equations:

1. \( x = 36z \)
2. \( y = 36z^2 + 5 \)

we can follow these steps:

### Step 1: Express \( z \) in terms of \( x \)

From the first equation, we can isolate \( z \):

\[ z = \frac{x}{36} \]

### Step 2: Substitute \( z \) into the equation for \( y \)

Now, we substitute \( z \) into the second equation:

\[ y = 36z^2 + 5 \]

Substituting \( z = \frac{x}{36} \):

\[ y = 36 \left( \frac{x}{36} \right)^2 + 5 \]

### Step 3: Simplify the equation

Now, we simplify the equation:

\[ y = 36 \left( \frac{x^2}{36^2} \right) + 5 \]

This simplifies to:

\[ y = 36 \cdot \frac{x^2}{1296} + 5 \]

Since \( 36 \div 1296 = \frac{1}{36} \):

\[ y = \frac{x^2}{36} + 5 \]

### Step 4: Write the final expression for \( y \)

Thus, we can express \( y \) in terms of \( x \):

\[ y = \frac{1}{36} x^2 + 5 \]

### Conclusion

The final expression for \( y \) in terms of \( x \) is:

\[ y = \frac{1}{36} x^2 + 5 \]