A student is scanning his 10 inch $\times$ 10 inch certificate at 600 dots per inch (dpi) to convert it to raster. What is the percentage reduction in number of pixels if the same certificate is scanned at 300 dpi?
Step 1: Calculate the number of pixels at 600 dpi.
- Resolution = 600 dpi $\Rightarrow$ in 10 inches, number of dots = $10 \times 600 = 6000$.
- So, image size = $6000 \times 6000 = 36,000,000$ pixels.
Step 2: Calculate the number of pixels at 300 dpi.
- Resolution = 300 dpi $\Rightarrow$ in 10 inches, number of dots = $10 \times 300 = 3000$.
- So, image size = $3000 \times 3000 = 9,000,000$ pixels.
Step 3: Percentage reduction.
\[
\text{Reduction} = \frac{\text{Initial pixels} - \text{Final pixels}}{\text{Initial pixels}} \times 100
\]
\[
= \frac{36,000,000 - 9,000,000}{36,000,000} \times 100
= \frac{27,000,000}{36,000,000} \times 100
= 75%
\]
\[
\boxed{75%}
\]
A function, \( \lambda \), is defined by \[ \lambda ( p,q ) = \begin{cases} (p - q)^2, & \text{if } p \geq q, \\ p + q, & \text{if } p < q. \end{cases} \] The value of the expression \( \dfrac{\lambda ( -(-3 + 2), (-2 + 3) )}{( -(-2 + 1) )} \) is:
Given two operators \( \oplus \) and \( \odot \) on numbers \( p \text{ and } q \) such that \[ p \oplus q = \frac{p^2 + q^2}{pq} \text{and} p \odot q = \frac{p^2}{q}, \] if \( x \oplus y = 2 \odot 2 \), then \( x = \)
Consider a five-digit number PQRST that has distinct digits P, Q, R, S, and T, and satisfies the following conditions:
1. \( P<Q \)
2. \( S>P>T \)
3. \( R<T \)
If integers 1 through 5 are used to construct such a number, the value of P is:



