To make the expression \(16a^2 – 12a\) a perfect square, we can complete the square by adding a term to it.

First, let’s factor out the common factor of 4:

\[16a^2 – 12a = 4(4a^2 – 3a)\]

Now, we want to complete the square for the expression inside the parentheses. The general form for a perfect square is \((x – y)^2 = x^2 – 2xy + y^2\). In our case, we have \(4a^2 – 3a\), so we can compare it to \(x^2 – 2xy\) to find the missing \(y^2\) term:

Comparing \(4a^2 – 3a\) to \(x^2 – 2xy\), we get:

– \(x = 2a\) (since \(x^2 = 4a^2\))
– \(-2xy = -3a\), so \(y = \frac{3}{4}\) (since \(x = 2a\))

Therefore, the missing \(y^2\) term is \(\left(\frac{3}{4}\right)^2 = \frac{9}{16}\).

However, remember that we factored out a 4 earlier, so we need to multiply this term by 4 to add it to the original expression:

\[4 \times \frac{9}{16} = \frac{9}{4}\]

So, the number that must be added to the expression \(16a^2 – 12a\) to make it a perfect square is \(\frac{9}{4}\).

Let’s denote the number of buffaloes as \(b\) and the number of ducks as \(d\).

Since each buffalo has 4 legs and each duck has 2 legs, the total number of legs in the group is \(4b + 2d\). The total number of heads, which is also the total number of animals, is \(b + d\).

According to the problem, the number of legs is 24 more than twice the number of heads. Therefore, we can write the equation:

\[4b + 2d = 2(b + d) + 24\]

Simplifying this equation:

\[4b + 2d = 2b + 2d + 24\]
\[2b = 24\]
\[b = 12\]

So, there are 12 buffaloes in the group.

The equation \(|x – 2| = 5\) can be solved by considering the two possible cases for the absolute value:

1. When \(x – 2 \geq 0\):
\[x – 2 = 5\]
\[x = 7\]

2. When \(x – 2 < 0\):
\[-(x – 2) = 5\]
\[x – 2 = -5\]
\[x = -3\]

Therefore, the solutions of the equation \(|x – 2| = 5\) are \(x = 7\) and \(x = -3\), which corresponds to option (b) \(-3, 7\).

To find the value of the Roman numeral MCDLXIV, we can break it down into its components:

– M = 1000
– CD = 400 (500 – 100)
– LX = 60 (50 + 10)
– IV = 4 (5 – 1)

Adding these values together, we get:

1000 + 400 + 60 + 4 = 1464

Therefore, the value of the numeral MCDLXIV is 1464, which corresponds to option (c) 1464.

Given the equation:
\[7 \sin^2 \theta + 3 \cos^2 \theta = 4\]

Since \(\cos^2 \theta = 1 – \sin^2 \theta\), we can substitute this into the equation:
\[7 \sin^2 \theta + 3 (1 – \sin^2 \theta) = 4\]

Expanding and simplifying:
\[7 \sin^2 \theta + 3 – 3 \sin^2 \theta = 4\]
\[4 \sin^2 \theta = 1\]
\[\sin^2 \theta = \frac{1}{4}\]
\[\sin \theta = \pm \frac{1}{2}\]

Since \(0^\circ < \theta < 90^\circ\), we know that \(\sin \theta\) is positive in this interval. Therefore, we can disregard the negative solution, leaving us with:
\[\sin \theta = \frac{1}{2}\]

The value of \(\theta\) that satisfies this equation in the given interval is:
\[\theta = 30^\circ \text{ or } \frac{\pi}{6} \text{ radians}\]

So, the value of \(\theta\) is \(30^\circ\) or \(\frac{\pi}{6}\) radians.