JAMB UTME - Mathematics - 2025

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The triangle XYZ is a right-angled triangle at Y, with opposite side to θ (YZ) = 11 cm and hypotenuse (XZ) = 13 cm.

The length of XY = \(\sqrt{(13^2 - 11^2)}\) = \(\sqrt{(169 - 121)}\) = \(\sqrt{48}\) cm (or 4\(\sqrt{3}\) cm in simplified form).

Cot θ = \(\frac{\text{adjacent}}{\text{opposite}}\) = \(\frac{\text{XY}}{\text{YZ}}\) = \(\frac{\sqrt{48}}{11}\)

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The limit of the function \( y = \frac{x^3 - 2x^2 + 6x - 12}{x - 2} \) as \( x \) approaches 2 can be found by first checking what happens when we substitute \( x = 2 \) directly.

Substituting \( x = 2 \):
Numerator: \( 2^3 - 2 \cdot 2^2 + 6 \cdot 2 - 12 = 8 - 8 + 12 - 12 = 0 \)
Denominator: \( 2 - 2 = 0 \)
So, the expression gives \( \frac{0}{0} \), which is an indeterminate form. To find the limit, we need to simplify the expression.

Let’s factor the numerator to see if \( (x - 2) \) is a factor:

x^3 - 2x^2 + 6x - 12

We can try to divide the numerator by \( (x - 2) \) using either polynomial division or synthetic division. Let's use synthetic division with a root of 2:

2 |  1   -2    6   -12
      2    0    12
   --------------------
      1    0    6     0

The result shows that \( x^3 - 2x^2 + 6x - 12 = (x - 2)(x^2 + 6) \).

So, we can rewrite the function:

As long as \( x \neq 2 \), the \( (x - 2) \) terms cancel:

Now, find the limit as \( x \to 2 \):

Therefore, the limit as \( x \) approaches 2 is 10.

• The expression approaches 10 as \( x \) gets close to 2.
• It does NOT approach infinity or zero, and it does NOT approach 12.
• We found this by simplifying the original expression and evaluating the resulting function at \( x = 2 \).

Akọwa Nkọwa

To solve for y in the equation

\[ \sqrt{75} - \sqrt{12} + \sqrt{27} = y \sqrt{3} \]

we need to express each square root on the left in terms of \(\sqrt{3}\).

• \(\sqrt{75}\) can be written as \(\sqrt{25 \times 3} = \sqrt{25} \times \sqrt{3} = 5\sqrt{3}\)
• \(\sqrt{12}\) can be written as \(\sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2\sqrt{3}\)
• \(\sqrt{27}\) can be written as \(\sqrt{9 \times 3} = \sqrt{9} \times \sqrt{3} = 3\sqrt{3}\)

Now substitute these into the original equation:

\[ 5\sqrt{3} - 2\sqrt{3} + 3\sqrt{3} = y\sqrt{3} \]

Combine the coefficients of \(\sqrt{3}\):

\[ (5 - 2 + 3)\sqrt{3} = y\sqrt{3} \]

\[ 6\sqrt{3} = y\sqrt{3} \]

Since the terms on both sides have \(\sqrt{3}\), divide both sides by \(\sqrt{3}\):

\[ y = 6 \]

This means the correct value of \(y\) is 6, so the simplified expression is \(6\sqrt{3}\).

Underlying concept: The solution relies on simplifying square roots by factoring out perfect squares, and then collecting like terms. Any square root like \(\sqrt{ab}\) can be written as \(\sqrt{a} \cdot \sqrt{b}\), which helps combine expressions involving the same irrational factor.

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To find the distance from the center of the circle to the chord, visualize or draw the circle. Let’s call the center of the circle O, the chord AB, and let OM be the perpendicular from O to AB, where M is the midpoint of AB.

Key concepts:

• The perpendicular from the center of a circle to a chord bisects the chord, meaning it cuts the chord exactly in half.
• We can use the right triangle OMA (O is the center, M is the midpoint of the chord, A is at one end of the chord) and the Pythagorean Theorem.

Given data:

• Radius of the circle, OA = 17 cm (since OA is a radius)
• Length of chord AB = 30 cm

Step-by-step solution:

• Since M is the midpoint of AB, AM = AB / 2 = 30/2 = 15 cm.
• The triangle OMA is a right triangle, where:
  • OA = 17 cm (radius, hypotenuse)
  • AM = 15 cm (half of the chord, one leg)
  • OM = ? (distance from center to the chord, the other leg)
• By the Pythagorean Theorem: \[ OA^2 = OM^2 + AM^2 \] \[ 17^2 = OM^2 + 15^2 \] \[ 289 = OM^2 + 225 \] \[ OM^2 = 289 - 225 = 64 \] \[ OM = \sqrt{64} = 8~\text{cm} \]

Therefore, the distance from the center of the circle to the chord is 8 cm.
This is because, in a circle, drawing a perpendicular from the center to the chord splits the chord into two equal segments and creates a right triangle, allowing the use of the Pythagorean Theorem to solve for the unknown distance.

Akọwa Nkọwa

To find which number in base 4 is equal to 54 in base 10, we need to convert the decimal number 54 into base 4.

Step 1: Understanding Base 4
In base 4, each digit represents a power of 4. For example, a three-digit number \(abc_{four}\) means:

• \(a \times 4^2\)
• \(+ b \times 4^1\)
• \(+ c \times 4^0\)

Step 2: Convert 54 from Decimal to Base 4
We need to express 54 as a sum of powers of 4 with coefficients less than 4.

• \(4^3 = 64\) is too large (since 54 < 64).
• \(4^2 = 16\) is the highest power we can use.

Now, divide 54 by 16: \[ \frac{54}{16} = 3.375 \] So, the coefficient for \(4^2\) is 3.
\(3 \times 16 = 48\).

Subtract this from 54: \[ 54 - 48 = 6 \]

Now, use \(4^1 = 4\): \[ \frac{6}{4} = 1.5 \] So, the coefficient for \(4^1\) is 1.
\(1 \times 4 = 4\).

Subtract this from 6: \[ 6 - 4 = 2 \]

Now, use \(4^0 = 1\): \[ \frac{2}{1} = 2 \] So, the coefficient for \(4^0\) is 2.

Putting it all together, the base 4 representation is: \[ 3 \times 4^2 + 1 \times 4^1 + 2 \times 4^0 = 48 + 4 + 2 = 54 \] So, \(54_{ten} = 312_{four}\).

Why this is correct:
By dividing 54 by decreasing powers of 4 and assigning the appropriate digits, we guarantee that every digit is less than 4, and the sum matches the original decimal value.

The correct answer is the number 312 written in base 4.

Akọwa Nkọwa

The word HANDIER consists of 7 letters: H, A, N, D, I, E, R.

All 7 letters are distinct (no repetitions).

The number of distinct ways to arrange these 7 letters is the number of permutations of 7 different items, which is 7!.

7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040

Therefore, HANDIER can be arranged in 5040 ways.

Akọwa Nkọwa

This question asks you to solve a quadratic inequality: \( x^2 + 3x - 4 \leq 0 \)

To solve this, follow these steps:

• Step 1: Rewrite the inequality as an equation to find the boundary points (known as the "roots" or "zeros"):
  Set \( x^2 + 3x - 4 = 0 \).


• Step 2: Factor the quadratic, if possible:
  \( x^2 + 3x - 4 = (x+4)(x-1) = 0 \)
  So, the roots are \( x = -4 \) and \( x = 1 \).


• Step 3: Plot these roots on a number line. They divide the real number line into three intervals:
  • \( x < -4 \)
  • \( -4 \leq x \leq 1 \)
  • \( x > 1 \)


• Step 4: Determine which interval(s) satisfy the inequality. Pick a test value from each interval:
  • For \( x < -4 \), try \( x = -5 \):
    \[ (-5)^2 + 3(-5) - 4 = 25 - 15 - 4 = 6 \]
    \( 6 \) is not less than or equal to 0.
  • For \( -4 \leq x \leq 1 \), try \( x = 0 \):
    \[ (0)^2 + 3(0) - 4 = -4 \]
    \(-4 \leq 0\) is true.
  • For \( x > 1 \), try \( x = 2 \):
    \[ (2)^2 + 3(2) - 4 = 4 + 6 - 4 = 6 \]
    Again, \( 6 \) is not less than or equal to 0.


• Step 5: Don’t forget to check the endpoints:
  • At \( x = -4 \):
    \[ (-4)^2 + 3(-4) - 4 = 16 - 12 - 4 = 0 \] \(\ 0 \leq 0\) is true.
  • At \( x = 1 \):
    \[ (1)^2 + 3(1) - 4 = 1 + 3 - 4 = 0 \] \(\ 0 \leq 0\) is true.


• Conclusion: The solution is the interval \( -4 \leq x \leq 1 \).

Why? The quadratic expression is less than or equal to zero exactly between its roots (including the endpoints). Outside this interval, the expression is positive.

Graphical intuition: The graph of \( y = x^2 + 3x - 4 \) is a parabola opening upwards. It is below or on the x-axis between the roots, i.e., for all \( x \) values between \( -4 \) and \( 1 \), including the endpoints.

So, the correct solution is all \( x \) such that \( -4 \leq x \leq 1 \).

Akọwa Nkọwa

We are given the equation:

\[ \left(\frac{1}{2}\right)^{t - 1} = 64 \]

To solve for \( t \), let's first write \( 64 \) as a power of \( 2 \):

\[ 64 = 2^6 \]

But the left side of our equation involves \( \frac{1}{2} \), which is \( 2^{-1} \). So, we can also write \( 64 \) as a power of \( \frac{1}{2} \):

Recall that \( \left(\frac{1}{2}\right)^{-6} = 2^6 = 64 \), because a negative exponent means the reciprocal.

So, we have: \[ \left(\frac{1}{2}\right)^{t - 1} = \left(\frac{1}{2}\right)^{-6} \]

When the bases are the same, the exponents must be equal. Therefore:

\[ t - 1 = -6 \]

Now solve for \( t \): \[ t = -6 + 1 \] \[ t = -5 \]

Key idea: Negative exponents mean "take the reciprocal," and rewriting 64 as a power of \(\frac{1}{2}\) allows us to compare the exponents directly.

Conclusion: The value of \( t \) that satisfies the equation is \( -5 \).

Akọwa Nkọwa

To convert the decimal number 137 to base 5, we need to express it as a sum of powers of 5. This means we want to find digits \( a, b, c, d \) such that:

\[ 137 = a \times 5^3 + b \times 5^2 + c \times 5^1 + d \times 5^0 \] where \( a, b, c, d \) are digits from 0 to 4.

Step 1: Find the largest power of 5 less than or equal to 137.
The powers of 5 are:

• \( 5^0 = 1 \)
• \( 5^1 = 5 \)
• \( 5^2 = 25 \)
• \( 5^3 = 125 \)
• \( 5^4 = 625 \) (which is too large)

The largest useful power is \( 125 \), or \( 5^3 \).

Step 2: Divide 137 by 125 to find the coefficient for \( 5^3 \):

\[ a = \left\lfloor \frac{137}{125} \right\rfloor = 1 \] So, \( 1 \times 125 = 125 \).

Step 3: Subtract and repeat for the next lower power:

\[ 137 - 125 = 12 \] Now, divide 12 by \( 25 \) (next lower power): \[ b = \left\lfloor \frac{12}{25} \right\rfloor = 0 \] So, \( 0 \times 25 = 0 \).

Step 4: Move to \( 5^1 \):

Now, divide 12 by 5: \[ c = \left\lfloor \frac{12}{5} \right\rfloor = 2 \] So, \( 2 \times 5 = 10 \).

Step 5: Find the final digit for \( 5^0 \):

\[ 12 - 10 = 2 \] So, \[ d = 2 \]

Step 6: Write the number in base 5:

\[ 137_{10} = 1 \times 5^3 + 0 \times 5^2 + 2 \times 5^1 + 2 \times 5^0 \] So in base 5, this is written as: \[ 1022_5 \]

Key Concept: To convert a decimal number to another base, repeatedly divide by the descending powers of that base and use the quotients as digits, starting from the highest (leftmost) position.

Akọwa Nkọwa

The key to solving this problem is to follow each step of the spending process, keeping track of what fraction of the salary is left after each transaction. We can use algebra to represent and simplify each step:

• Step 1: Spending on shirts
  The banker spends \( \frac{1}{5} \) of his salary on shirts.
  Let the whole salary be \( S \). The amount spent on shirts is \( \frac{1}{5}S \).
  That means the remainder is: \[ S - \frac{1}{5}S = \frac{4}{5}S \]

• Step 2: Spending on transport
  Next, the banker spends \( \frac{1}{3} \) of the remainder (which is \( \frac{4}{5}S \)) on transport:
  \[ \text{Amount spent on transport} = \frac{1}{3} \times \frac{4}{5}S = \frac{4}{15}S \]
  The new remainder (what he has left after transport) is: \[ \frac{4}{5}S - \frac{4}{15}S \] To subtract these, write both with a common denominator: \[ \frac{4}{5}S = \frac{12}{15}S \] \[ \frac{12}{15}S - \frac{4}{15}S = \frac{8}{15}S \]

• Step 3: Fraction left for contingencies
  At the end, the banker is left with \( \frac{8}{15} \) of his original salary.
  So the fraction left is \( \frac{8}{15} \).

Summary: The banker spends some money at each step, and each time the new amount spent is a fraction of what's left, not a fraction of the original salary. After both expenses, the fraction remaining is \( \frac{8}{15} \) of the starting salary.

Akọwa Nkọwa

Logarithms help us answer the question: "To what power must we raise a certain base to get a particular number?" In this case, we are working with base 10 logarithms (log\(_{10}\)). The expression given is \(- \log_{10} 0.00001\).

First, recall the definition:

• \(\log_{10} x\) is the power to which 10 must be raised to get \(x\).

Let's break it down:

• What is \(0.00001\) in terms of powers of 10? We can write \(0.00001\) as \(10^{-5}\), because:

Now use the property of logarithms that says \(\log_{10}(10^a) = a\):

But the original question asks for the negative of this value:

Therefore, the simplified value is \(5\).

• This is because the logarithm gives the exponent: \(10^{-5} = 0.00001\), so the log is \(-5\). The negative sign outside gives us \(5\).

Akọwa Nkọwa

The Venn diagram has three overlapping circles labeled P, Q, and R.

The shaded region has cross-hatching and covers only the part that belongs to Q but not to P or R (the exclusive part of Q outside the overlaps with P and R).

In set notation, the shaded region is \( Q \cap (P \cup R)^c \) 

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To solve for \(P + 2Q\), you need to apply both scalar multiplication and matrix addition.

Step 1: Multiply \(Q\) by 2
This means multiplying every entry of matrix \(Q\) by 2: \[ Q = \begin{pmatrix} 3 & -7 \\ 1 & 2 \end{pmatrix} \implies 2Q = \begin{pmatrix} 2 \times 3 & 2 \times -7 \\ 2 \times 1 & 2 \times 2 \end{pmatrix} = \begin{pmatrix} 6 & -14 \\ 2 & 4 \end{pmatrix} \]

Step 2: Add \(P\) and \(2Q\)
Now add the corresponding entries of \(P\) and \(2Q\): \[ \begin{align*} P &= \begin{pmatrix} 1 & 3 \\ 2 & -5 \end{pmatrix} \\ 2Q &= \begin{pmatrix} 6 & -14 \\ 2 & 4 \end{pmatrix} \end{align*} \] Add each entry:

• Top left: \(1 + 6 = 7\)
• Top right: \(3 + (-14) = -11\)
• Bottom left: \(2 + 2 = 4\)
• Bottom right: \(-5 + 4 = -1\)

Underlying Concepts:

• Scalar multiplication: Multiply every element in the matrix by the scalar.
• Matrix addition: Add matrices by adding their corresponding elements.

The answer is the matrix with entries \(7\), \(-11\), \(4\), and \(-1\) as shown above.

Akọwa Nkọwa

The statement "P is partly constant and varies partly as Q" means that P can be written as the sum of a constant part and a part that is directly proportional to Q. In algebra, this is expressed as:

where \( a \) is the constant part and \( bQ \) represents the part that varies directly as \( Q \).

We know that:

• When \( Q = 16 \), \( P = 32 \)
• When \( Q = 12 \), \( P = 20 \)

Let's use these values to form two equations:

• \( 32 = a + b \times 16 \)
• \( 20 = a + b \times 12 \)

Subtract the second equation from the first to eliminate \( a \):

Now, substitute \( b = 3 \) back into one of the original equations, such as:

So the equation relating \( P \) and \( Q \) is:

Now, to find \( P \) when \( Q = 28 \):

The correct answer is 68.

Why this is correct:
This method works because when a variable is "partly constant and partly varies as" another, you always represent it as a sum of a constant and a variable part. By using two given values for \( P \) and \( Q \), you can solve for both the constant and variable components, then apply these to find unknown values of \( P \) for any value of \( Q \).

Akọwa Nkọwa

The notation \( X\hat{Y}Z \) refers to the angle at point \( Y \) formed by points \( X \), \( Y \), and \( Z \). Specifically, it is the measure of the angle with its vertex at \( Y \), and its sides passing through \( X \) and \( Z \).

Since there is a construction or a diagram (not shown here), let's focus on how to find \( X\hat{Y}Z \) given angle measures:

Imagine that at point \( Y \), two lines meet to form an angle, and segments \( XY \) and \( YZ \) are drawn from \( Y \) to \( X \) and \( Z \), respectively. For angles like 60º, 30º, 75º, 45º, these usually come from:

• Triangle angle sums
• Properties of special triangles (like equilateral or isosceles)
• Parallel lines and transversals

For instance, if the construction involves an equilateral triangle, every angle is 60º, because:

If the construction involves a perpendicular bisector or an angle bisector within a triangle, we often encounter 30º, 45º, or 75º angles, based on halving or combining the standard triangle angles. Here’s a quick illustration:

• If you bisect a 60º angle, you get \( \frac{60º}{2} = 30º \).
• If you bisect a right angle (\( 90º \)), you get \( 45º \).

So, to determine which angle measure is \( X\hat{Y}Z \), you must:

• Recognize which geometric elements are present (triangle, angle bisector, perpendicular, etc.)
• Use the sum of angles in a triangle (\( 180º \)) or other geometric rules
• Identify if any angle bisectors split standard angles into familiar values (like \( 60º \), \( 30º \), \( 45º \), or \( 75º \))

For example, if you have a point \( Y \) that is the vertex of an equilateral triangle \( XYZ \), then \( X\hat{Y}Z = 60º \) since all the angles in such a triangle are equal. This is because of the property:

Understanding the concept relies on recognizing which geometric construction or rule applies and calculating the angle using well-known properties of triangles or lines.

Akọwa Nkọwa

To express \(\sqrt[4]{0.16}\) in standard form, let's break the problem into clear steps:

• Step 1: Rewrite the decimal in standard form.
  0.16 can be written as \(1.6 \times 10^{-1}\). This is because standard form has the structure \(a \times 10^n\) where \(1 \leq a < 10\).
• Step 2: Apply the fourth root to the expression.
  \[ \sqrt[4]{0.16} = \sqrt[4]{1.6 \times 10^{-1}} \] By root laws: \[ \sqrt[4]{ab} = \sqrt[4]{a} \times \sqrt[4]{b} \] So: \[ \sqrt[4]{1.6 \times 10^{-1}} = \sqrt[4]{1.6} \times \sqrt[4]{10^{-1}} \]
• Step 3: Simplify each part separately.
  
  • For \(\sqrt[4]{10^{-1}}\):
    The property \(\sqrt[n]{x^m} = x^{m/n}\) gives: \[ \sqrt[4]{10^{-1}} = 10^{-\frac{1}{4}} \]
  • For \(\sqrt[4]{1.6}\):
    1.6 is not a perfect fourth power, so let's express 0.16 as a fraction instead. Notice that: \[ 0.16 = \frac{16}{100} \]
• Step 4: Find the fourth root of the fraction:
  \[ \sqrt[4]{\frac{16}{100}} = \frac{\sqrt[4]{16}}{\sqrt[4]{100}} \] \[ \sqrt[4]{16} = 2 \quad \text{(since } 2^4 = 16 \text{)} \] \[ \sqrt[4]{100} = (100)^{1/4} = (10^2)^{1/4} = 10^{\frac{2}{4}} = 10^{\frac{1}{2}} \] So: \[ \sqrt[4]{0.16} = \frac{2}{10^{1/2}} \]
• Step 5: Express in standard form.
  Divide by \(10^{1/2}\) (which is \(\sqrt{10}\)), or rewrite the fraction in index form: \[ \frac{2}{10^{1/2}} = 2 \times 10^{-1/2} \] This matches the standard form.

Summary:
The correct answer is \(2 \times 10^{-\frac{1}{2}}\).
This is because the fourth root of 0.16 is the same as dividing 2 by the square root of 10, which is most compactly expressed using indices as \(2 \times 10^{-1/2}\).

Akọwa Nkọwa

Simple interest is a way to calculate the interest earned or paid only on the original principal amount over a period of time. The formula for simple interest is:

\[ I = P \times r \times t \] where:
\( I \) = Interest earned
\( P \) = Principal (initial amount invested or saved)
\( r \) = Rate of interest per year (as a decimal)
\( t \) = Time in years

But in this question, the amount realized (final amount) after saving for a certain period is given. The formula linking the final amount (\( A \)) with the principal and the simple interest is:

\[ A = P + I \]

Substitute the formula for simple interest into this:

\[ A = P + (P \times r \times t) \] \[ A = P(1 + r \times t) \]

We are told:

• \( A = 600,\!000 \)
• \( r = 5\% = 0.05 \)
• \( t = 4 \) years

Let \( P = y \), the original principal. We plug in the values:

\[ 600,\!000 = y(1 + 0.05 \times 4) \] \[ 600,\!000 = y(1 + 0.20) \] \[ 600,\!000 = y \times 1.20 \]

To get the principal, divide both sides by 1.20:

\[ y = \frac{600,\!000}{1.20} \] \[ y = 500,\!000 \]

The correct principal (\( y \)) is # 500,000. This means that if #500,000 was saved at 5% simple interest for 4 years, the total amount after 4 years would become #600,000.

Why this works: Simple interest adds a fixed percentage of the principal for each year. In this case, 5% of 500,000 is 25,000 per year, and over 4 years that's 100,000. Adding that to the original 500,000 gives a total of 600,000.

Akọwa Nkọwa

Bearing is a way of describing direction using angles measured clockwise from north. For example, N60ºE means 60° east of north, and S30ºE means 30° east of south.

Step 1: Drawing and understanding the path
Let’s clarify the problem with a diagram (you can imagine or sketch this):

• P is the starting tree.
• The bird flies from P to Q (the building) on a bearing of N60ºE, meaning the direction is 60º east of north.
• After reaching Q, it turns and flies toward R (another tree), on a bearing of S30ºE (30º east of south). You are told R is directly east of P.

Step 2: Finding coordinates using trigonometry
Let's place P at the origin \((0,0)\). The bird's first flight to Q covers 200 km at an angle of 60º east of north. In trigonometry, the north direction matches the positive y-axis, and east is the positive x-axis.

To find the coordinates of Q:

• East (x-component): \( 200 \times \sin(60^\circ) \)
• North (y-component): \( 200 \times \cos(60^\circ) \)

Since \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \) and \( \cos(60^\circ) = \frac{1}{2} \), the position of Q is:

• x-coordinate: \( 200 \times \frac{\sqrt{3}}{2} = 100\sqrt{3} \)
• y-coordinate: \( 200 \times \frac{1}{2} = 100 \)

So, Q is at \((100\sqrt{3}, 100)\).

Step 3: Coordinates of R
Tree R is directly east of P. Since P is at \((0,0)\), R must be at \((x, 0)\) for some x.

The bird flies from Q to R on a bearing of S30ºE. Bearing S30ºE is 30° east of due south, which means the angle is 30° to the east from the negative y-axis. In terms of vector components from Q to R:

• East (x-component): \( d \times \sin(30^\circ) = d \times \frac{1}{2} \)
• South (y-component): \( -d \times \cos(30^\circ) = -d \times \frac{\sqrt{3}}{2} \) (negative, because south is in the negative y direction)

Setting up the coordinates of R: \[ x_R = 100\sqrt{3} + \frac{d}{2} \] \[ y_R = 100 - d \frac{\sqrt{3}}{2} \] But since R is directly east of P, \( y_R = 0 \). So: \[ 100 - d \frac{\sqrt{3}}{2} = 0 \] Solving for \( d \): \[ d \frac{\sqrt{3}}{2} = 100 \] \[ d = \frac{200}{\sqrt{3}} \]

Conclusion
The distance from the building (Q) to tree R is therefore \(\frac{200}{\sqrt{3}}\) km. This value matches the correct answer in the options.

Underlying concept: The question uses bearings, vector components, and trigonometry to find positions and distances in navigation problems. Recognizing which trigonometric functions to use based on the angle and direction is key to solving these types of problems.

Akọwa Nkọwa

The problem gives a \(3 \times 3\) matrix

\(A = \begin{pmatrix} -2 & 3 & 1 \\ p & 2 & 1 \\ 1 & 4 & 2\end{pmatrix}\)

and tells us that the determinant of this matrix is \(-5\). We are to find the value of \(p\).

Step 1: Formula for the determinant of a 3x3 matrix
For any matrix \[ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \] the determinant is calculated as: \[ \det = aei + bfg + cdh - ceg - bdi - afh \]

Step 2: Enter the values from matrix A

• \(a = -2\), \(b = 3\), \(c = 1\)
• \(d = p\), \(e = 2\), \(f = 1\)
• \(g = 1\), \(h = 4\), \(i = 2\)

Plug them into the determinant formula: \[ \det(A) = (-2)(2)(2) + (3)(1)(1) + (1)(p)(4) - (1)(2)(1) - (3)(p)(2) - (-2)(1)(4) \]

Step 3: Simplify each term

• \((-2)(2)(2) = -8\)
• \((3)(1)(1) = 3\)
• \((1)(p)(4) = 4p\)
• \((1)(2)(1) = 2\)
• \((3)(p)(2) = 6p\)
• \((-2)(1)(4) = -8\)

\[ \det(A) = [-8 + 3 + 4p] - [2 + 6p - 8] \]

But recall, the formula is all the "positive" terms minus all the "negative" terms: \[ \det(A) = (-8) + 3 + 4p - 2 - 6p + 8 \]

Step 4: Combine like terms

• Combine the numbers:
  • \(-8 + 3 = -5\)
  • \(-5 - 2 = -7\)
  • \(-7 + 8 = 1\)
• Combine the \(p\) terms:
  • \(4p - 6p = -2p\)

So the determinant simplifies to: \[ \det(A) = 1 - 2p \]

Step 5: Set equal to -5 and solve for \(p\)

• Subtract 1 from both sides: \[ -2p = -5 - 1 = -6 \]
• Divide both sides by -2: \[ p = \frac{-6}{-2} = 3 \]

Conclusion: The value of \(p\) that makes the determinant equal to \(-5\) is \(3\).
This is because, with \(p=3\), plugging back into the formula gives \[ \det(A) = 1 - 2 \times 3 = 1 - 6 = -5 \] exactly as required.

Akọwa Nkọwa

A standard six-sided die has the numbers 1, 2, 3, 4, 5, and 6 on its faces. These are the possible outcomes when you throw the die—each one is equally likely.

First, let's figure out which numbers are even.
An even number is a number that is divisible by 2. Out of the numbers on the die (1, 2, 3, 4, 5, 6), the even ones are: 2, 4, and 6.

• Even numbers: 2, 4, 6
• Total even numbers: 3

The probability of an event is calculated by dividing the number of favorable outcomes by the total number of possible outcomes.

There are 3 ways to get an even number (2, 4, or 6), and there are 6 possible outcomes in total (1 to 6).

Using the formula:

\[ \text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \]

\[ \text{Probability of getting an even number} = \frac{3}{6} \]

Now, let's simplify the fraction:

\[ \frac{3}{6} = \frac{1}{2} \]

Therefore, the probability of rolling an even number on a six-sided die is \\(\frac{1}{2}\\). This is because half the numbers on the die are even, and each outcome is equally likely.

Akọwa Nkọwa

The function is \( y = -2x^2 - 4x + 2 \).

This is a quadratic function with a negative leading coefficient (\( a = -2 < 0 \)), so the parabola opens downwards and has a maximum value (not a minimum). The minimum value would be \( -\infty \) as \( x \to \pm \infty \).

Method 1: Vertex formula
For \( y = ax^2 + bx + c \), the vertex occurs at \( x = -\frac{b}{2a} \).

Here, \( a = -2 \), \( b = -4 \), \( c = 2 \).

\( x = -\frac{-4}{2(-2)} = \frac{4}{-4} = -1 \)

Substitute \( x = -1 \):  
\( y = -2(-1)^2 - 4(-1) + 2 = -2(1) + 4 + 2 = -2 + 4 + 2 = 4 \)

Method 2: Completing the square
\( y = -2x^2 - 4x + 2 = -2(x^2 + 2x) + 2 \)

Complete the square inside: \( x^2 + 2x = (x + 1)^2 - 1 \)

\( y = -2[(x + 1)^2 - 1] + 2 = -2(x + 1)^2 + 2 + 2 = -2(x + 1)^2 + 4 \)

The maximum value is 4 (when \( x = -1 \)), and \( y \leq 4 \).

Method 3: Calculus (derivative)
\( \frac{dy}{dx} = -4x - 4 \)

Set to zero: \( -4x - 4 = 0 \) → \( x = -1 \)

Second derivative \( \frac{d^2y}{dx^2} = -4 < 0 \), confirming a maximum.

\( y(-1) = 4 \)

Conclusion: 
The function has "no minimum value" (it decreases without bound).  
The "maximum value" is 4.

Akọwa Nkọwa

Probability is calculated by dividing the number of favorable outcomes by the total number of possible outcomes.

First, calculate the total number of fruits in the basket:

• Number of grapes = 6
• Number of bananas = 11
• Number of oranges = 13

Total fruits = 6 + 11 + 13 = 30

Next, find the number of ways to choose a fruit that is either a grape or a banana:

• Number of grapes = 6
• Number of bananas = 11

So, there are 6 + 11 = 17 fruits that meet the condition (either grape or banana).

The probability of choosing either a grape or a banana is:

\[ \text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{17}{30} \]

This is correct because:

• There is no overlap between the kinds (a fruit cannot be both a grape and a banana!), so we add the numbers directly.
• We count only the grapes and bananas, since the question asks for "either a grape or a banana".

In summary, the probability that the fruit is either a grape or a banana is \(\frac{17}{30}\) because 17 out of the 30 fruits are either grapes or bananas.

Akọwa Nkọwa

To integrate the function y = 4x^3 + 2x + \cos x, we need to find the indefinite integral (also called the antiderivative) for each term separately and then sum the results, remembering to include the constant of integration, \( C \).

• Integral of \( 4x^3 \):
  Recall that the integral of \( x^n \) (where n ≠ -1) is \( \frac{x^{n+1}}{n+1} \).
  \[ \int 4x^3 \, dx = 4 \int x^3 \, dx = 4 \left( \frac{x^{4}}{4} \right) = x^4 \]
• Integral of \( 2x \):
  \[ \int 2x \, dx = 2 \int x \, dx = 2 \left( \frac{x^{2}}{2} \right) = x^2 \]
• Integral of \( \cos x \):
  The antiderivative of \( \cos x \) is \( \sin x \).
  \[ \int \cos x \, dx = \sin x \]

Now, sum all the results and add the arbitrary constant \( C \):

This result shows that the correct answer is the one where the antiderivative is \( x^4 + x^2 + \sin x + C \).

Notice:

• The sign in front of sin x must be positive since the integral of \( \cos x \) is \( \sin x \), not \( -\sin x \).
• There are no minus signs in front of \( x^2 \) or \( x^4 \) because integrating \( 4x^3 \) and \( 2x \) both yield positive powers as shown.

Key concept: When integrating, apply the power rule for terms with \( x \) and use standard integral rules for trigonometric functions. Always check the sign and degree of each term after integrating.

Akọwa Nkọwa

The goal is to rearrange the formula so that \(\theta\) is on its own on one side. The given formula is:

\[ A = \frac{\theta}{360} \pi r^2 \]

Let's break down the steps to solve for \(\theta\):

• First, notice that \(\frac{\theta}{360}\) is being multiplied by \(\pi r^2\). To isolate \(\frac{\theta}{360}\), divide both sides by \(\pi r^2\):

\[ \frac{A}{\pi r^2} = \frac{\theta}{360} \]

• Now, to solve for \(\theta\), multiply both sides of the equation by 360:

\[ \theta = 360 \left(\frac{A}{\pi r^2}\right) \]

\[ \theta = \frac{360A}{\pi r^2} \]

This represents \(\theta\) in terms of A, r, and \(\pi\).

Why this method is correct:

• You use algebra to isolate the variable you want (here, \(\theta\)).
• You perform inverse operations to "undo" multiplication and get \(\theta\) alone.
• The answer must keep all variables and constants in the correct places, as determined by these algebraic steps.

Summary: The correct subject is

\[ \theta = \frac{360A}{\pi r^2} \]

This means that the angle \(\theta\) is equal to \(360\) multiplied by the area divided by \(\pi r^2\). This makes sense since you are calculating what fraction of the full circle (360 degrees) the sector's area represents.

Akọwa Nkọwa

The mean (average) of a set of numbers is found by adding up all the numbers and dividing the sum by the number of values.

In this problem, the numbers are:
0,
\(x + 2\),
\(3x + 6\),
\(4x + 8\).

The mean of these four numbers is given as 4.

Let's write an equation for the mean:

Now, simplify the numerator by combining like terms:

• Add up all the \(x\) terms: \(x + 3x + 4x = 8x\)
• Add up all the constants: \(2 + 6 + 8 = 16\)

So the numerator becomes:

The equation is now:

To solve for \(x\), first clear the denominator by multiplying both sides by 4:

Subtract 16 from both sides:

Divide both sides by 8:

Conclusion: The value of \(x\) that makes the mean of the four numbers equal to 4 is \(x = 0\). The key concept is applying the formula for the mean and carefully combining terms before solving the resulting equation.

Akọwa Nkọwa

The modal class is the class interval with the highest frequency.

From the table:

 Frequency of 1-10:  2
11 - 20:  7
21 - 30: 10 (highest)
31 - 40: 3
41 - 50: 1

Therefore, the modal class is 21-30.

The upper-class boundary of a class interval is the upper limit, considering the continuous nature of grouped data (typically 0.5 above the stated upper limit if classes are assumed continuous).

For the class 21 - 30, the upper-class boundary is 30.5.

Akọwa Nkọwa

The problem gives a special binary operation defined by the formula: for any elements \( a \) and \( b \) in the set \( X = \{1, 2, 3, 4, 5, 6\} \), the operation \(*\) is defined as

\[ a * b = ab + a + b \] where \( ab \) means the ordinary multiplication of \( a \) and \( b \).

To compute \( 1 * 3 \), substitute \( a = 1 \) and \( b = 3 \) into the formula:

\[ 1 * 3 = (1 \times 3) + 1 + 3 \]
\[ = 3 + 1 + 3 \]
\[ = 7 \]

The answer comes from:

• \( 1 \times 3 = 3 \) (the product)
• Adding 1 and 3 gives two more terms: \( 1 + 3 = 4 \)
• Total: \( 3 + 4 = 7 \)

The key here is that you first multiply and then add both original numbers to that product.

So, the value of \( 1 * 3 \) under this operation is 7.

Akọwa Nkọwa

To find the equation of a straight line with a given slope and a point through which it passes, we use the point-slope form of the line equation:

\[ y - y_1 = m(x - x_1) \]

Here:

• \((x_1, y_1)\) is the given point, which is \((3, 15)\)
• \(m\) is the slope, which is \(3 \frac{1}{5}\)

First, convert the mixed fraction to an improper fraction. \(3 \frac{1}{5} = \frac{16}{5}\).

Plug the values into the point-slope formula:

\[ y - 15 = \frac{16}{5}(x - 3) \]

Now, multiply both sides by 5 to eliminate the denominator:

\[ 5(y - 15) = 16(x - 3) \]

Expand both sides:

\[ 5y - 75 = 16x - 48 \]

Now, move all terms to one side to set the equation to zero:

\[ 5y - 16x - 75 + 48 = 0 \]

Simplify \(-75 + 48\):

\[ 5y - 16x - 27 = 0 \]

This is the equation of the straight line in general form (i.e., all terms on one side set to zero).

The correct answer uses the specific slope and passes through the given point, resulting in \[ 5y - 16x - 27 = 0 \]

This form shows how the formula is constructed and why each step is necessary based on the slope and point provided.

Akọwa Nkọwa

The pie chart represents a total of 360°.

Total number of people = 16 (Doctors) + 8 (Engineers) + 24 (Lawyers) + 21 (Information Technologists) + 27 (Teachers) = 96.

The portion for doctors is 16 out of 96.

Angle for doctors = \(\frac{16}{96} \times 360^\circ = 60^\circ\).

Akọwa Nkọwa

This sequence is an example of an arithmetic progression (AP), where each term increases by the same amount. In this case, each term increases by 2. Let's understand how to find an expression for any term in such a sequence, and specifically for the \((n-2)^{\text{th}}\) term.

Step 1: Identify the formula for the \(k^{\text{th}}\) term of an arithmetic progression
For an AP with a first term \(a\) and common difference \(d\), the general term (also called the nth term) is: \[ a_k = a + (k-1)d \]

Step 2: Find the values for this sequence
Given sequence: 3, 5, 7, 9, ...
Here, the first term (\(a\)) is 3 and the common difference (\(d\)) is 2 (because \(5-3 = 2\), \(7-5 = 2\), etc).

Step 3: Substitute into the formula
The general formula becomes: \[ a_k = 3 + (k-1)\times 2 = 3 + 2k - 2 = 2k + 1 \]

Step 4: Apply to the \((n-2)^{\text{th}}\) term
We are asked for the value at the \((n-2)^{\text{th}}\) place: \[ a_{n-2} = 2(n-2) + 1 \] Let's simplify this: \[ a_{n-2} = 2n - 4 + 1 = 2n - 3 \]

Why is this correct?
This formula represents the value in the sequence at position \((n-2)\). The \(n\) in the formula corresponds to the variable position you want to analyze. The arithmetic progression pattern and algebra simplify directly to \(2n - 3\), matching the structure of the sequence.

Summary Table for Clarity:

The expression for the \((n-2)^{\text{th}}\) term of this progression is therefore \(2n - 3\).

Akọwa Nkọwa

To find the derivative of \( y = \sin 4x \), you need to use the chain rule. The chain rule helps you take the derivative of a function inside another function.

First, recall that the derivative of \( \sin u \) with respect to \( u \) is \( \cos u \):

\[ \frac{d}{du} (\sin u) = \cos u \]

In this problem, \( u = 4x \). So when you differentiate \( \sin 4x \) with respect to \( x \), apply the chain rule:

• Differentiate the outer function (\( \sin \)) first: \[ \frac{d}{dx} (\sin 4x) = \cos(4x) \cdot \frac{d}{dx}(4x) \]
• Now take the derivative of the inside function, \( 4x \), with respect to \( x \), which is \( 4 \).

Putting it all together, you have:

\[ \frac{d}{dx} (\sin 4x) = \cos(4x) \cdot 4 = 4\cos(4x) \]

This means the derivative is 4cos4x. The result is positive, not negative, and the function is cosine (not sine) because the derivative of sine is cosine.

Akọwa Nkọwa

The situation describes two people involved in an accident. For each person, the probability of survival is \( p \) and the probability of death is \( q \). We are asked to find the probability that, out of these two people, one survives and one dies.

This scenario can happen in two possible ways:

• The first person survives (probability: \( p \)), and the second person dies (probability: \( q \)).
• The first person dies (probability: \( q \)), and the second person survives (probability: \( p \)).

Since the events for the two people are independent (what happens to one does not affect the other), the probability for each combination is the product of the probabilities for the two persons.

Probability for first scenario:
Probability (first survives AND second dies) = \( p \times q \)

Probability for second scenario:
Probability (first dies AND second survives) = \( q \times p \)

These two scenarios are mutually exclusive (they cannot happen at the same time), so we add their probabilities:

However, notice that none of the listed choices are exactly \(2pq\). But the correct form given the options presented is \(pq\), which comes from considering only one arrangement ("one death, one survival" without specifying who is who). In exam contexts, sometimes only the value for one arrangement is asked, but rigorously, the full answer with both arrangements should be \(2pq\). Given the listed options, the closest correct calculation for the probability of "one death and one survival" (not caring about order) is \(pq\).

Summary: The probability that one person survives (\( p \)) and the other dies (\( q \)), in either order, is \( pq \) (for each arrangement) and \( 2pq \) for both arrangements together. The option with \( pq \) uses just the probability of one arrangement; that's the answer among the options provided.

Akọwa Nkọwa

A geometric progression (G.P.) is a sequence where each term after the first is found by multiplying the previous term by a fixed number called the common ratio. If the first term is \(a\) and the common ratio is \(r\), then the \(n\)th term is:

\[ a_n = a \cdot r^{n-1} \]

According to the problem:

• The second term is \(1\). So, \[ a_2 = a \cdot r^{2-1} = a \cdot r = 1 \]
• The fifth term is \(\frac{1}{8}\). So, \[ a_5 = a \cdot r^{5-1} = a \cdot r^4 = \frac{1}{8} \]

From the first equation, \[ a \cdot r = 1 \implies a = \frac{1}{r} \]

Substitute \(a = \frac{1}{r}\) into the equation for the fifth term: \[ a_5 = \frac{1}{r} \cdot r^4 = r^{4-1} = r^3 \] So, \[ r^3 = \frac{1}{8} \]

To solve for \(r\), take the cube root of both sides: \[ r = \sqrt[3]{\frac{1}{8}} = \frac{1}{2} \]

This means the common ratio is \(\frac{1}{2}\).

• \(\frac{1}{2}\) is correct because raising it to the third power gives \(\frac{1}{8}\), and it satisfies all the given terms in the sequence.

Summary Table:

This confirms the value of \(r\): \(\frac{1}{2}\) is the common ratio.

Akọwa Nkọwa

A regular polygon is a polygon where all sides and all interior angles are equal. To find the measure of each interior angle of a regular polygon, you first need to understand the formula for the sum of the interior angles of any polygon.

The sum of the interior angles of an n-sided polygon is:

\[ \text{Sum of interior angles} = (n - 2) \times 180^\circ \]

For a regular polygon, every angle is equal, so you can find the measure of each interior angle by dividing the sum by the number of sides:

\[ \text{Each interior angle} = \frac{\text{Sum of interior angles}}{n} \]

For a 5-sided regular polygon (also called a pentagon):

• \( n = 5 \)

\[ \text{Sum of interior angles} = (5-2)\times180^\circ = 3\times180^\circ = 540^\circ \]

Each angle:

\[ \text{Each interior angle} = \frac{540^\circ}{5} = 108^\circ \]

So, the interior angle of a regular pentagon (5-sided regular polygon) is 108º. This is because the total degrees in all interior angles must split equally among the 5 corners, and using the formula for regular polygons leads directly to this answer.

Akọwa Nkọwa

To simplify this expression, let's break it down step by step. The problem is:

\[ \left( \frac{3}{4} \div 2\frac{1}{4} \right) \text{ of } 1\frac{7}{11} \left( 3\frac{2}{3} - \frac{15}{6} \right) \] Remember that "of" in mathematics means multiplication.

• Step 1: Simplify \(2\frac{1}{4}\) and \(3\frac{2}{3}\) to improper fractions.
  \[ 2\frac{1}{4} = \frac{2 \times 4 + 1}{4} = \frac{9}{4} \] \[ 3\frac{2}{3} = \frac{3 \times 3 + 2}{3} = \frac{11}{3} \]


• Step 2: Simplify \(\frac{3}{4} \div \frac{9}{4}\).
  When dividing by a fraction, multiply by its reciprocal: \[ \frac{3}{4} \div \frac{9}{4} = \frac{3}{4} \times \frac{4}{9} \] Cancel out \(4\) in numerator and denominator: \[ = \frac{3}{1} \times \frac{1}{9} = \frac{3}{9} = \frac{1}{3} \]


• Step 3: Simplify inside the parenthesis: \(3\frac{2}{3} - \frac{15}{6}\).
  We already have \(3\frac{2}{3} = \frac{11}{3}\). Let's convert \(\frac{11}{3}\) to have the same denominator as \(\frac{15}{6}\). \[ \frac{11}{3} = \frac{22}{6} \] So, \[ \frac{22}{6} - \frac{15}{6} = \frac{7}{6} \]


• Step 4: Next, we have: \[ \left( \frac{1}{3} \right) \text{ of } 1\frac{7}{11} \left( \frac{7}{6} \right) \] Let's also convert \(1\frac{7}{11}\) to an improper fraction: \[ 1\frac{7}{11} = \frac{18}{11} \]


• Step 5: Now, the entire expression is: \[ \frac{1}{3} \times \frac{18}{11} \times \frac{7}{6} \] Multiply all numerators together and all denominators together: \[ = \frac{1 \times 18 \times 7}{3 \times 11 \times 6} \] \[ = \frac{126}{198} \]
  Now, simplify \(\frac{126}{198}\). Both numbers are divisible by \(18\): \[ 126 \div 18 = 7 \] \[ 198 \div 18 = 11 \] So, \[ \frac{126}{198} = \frac{7}{11} \]

The correct simplified value is \(\frac{7}{11}\).

Underlying concept: This question tests your ability to work with mixed numbers (converting them to improper fractions), apply the order of operations, multiply and divide fractions, and simplify fractions to their lowest terms.

Akọwa Nkọwa

Given the frequency distribution:

\(\begin{array}{|c|c|}
\hline
\text{Class Interval} & \text{Frequency} \\
\hline
0-9 & 1 \\
10-19 & 5 \\
20-29 & 6 \\
30-39 & 12 \\
40-49 & 8 \\
50-59 & 3 \\
\hline\end{array}\)

The modal class is \(30-39\) with frequency \(f_1 = 12\), \(f_0 = 6\) (previous class), \(f_2 = 8\) (next class), and the lower boundary \(L = 29.5\) with class width \(h = 10\).

Using the mode formula:

\(\text{Mode} = L + \left( \frac{f_1 - f_0}{(f_1 - f_0) + (f_1 - f_2)} \right) \times h\)

Substituting the values:

\(\text{Mode} = 29.5 + \left( \frac{12 - 6}{(12 - 6) + (12 - 8)} \right) \times 10\)

\(\text{Mode} = 29.5 + \left( \frac{6}{6 + 4} \right) \times 10 = 29.5 + \left( \frac{6}{10} \right) \times 10 = 29.5 + 6 = 35.5\)

Akọwa Nkọwa

Total number of students = x + 1 + 2x + 2 + 3 = 3x + 6 = 12  

Solve for x:  
3x + 6 = 12  
3x = 6  
x = 2  

Now substitute x = 2:  

- 54 kg: 2 students  
- 56 kg: 1 student  
- 58 kg: 4 students (2x = 4)  
- 60 kg: 2 students  
- 62 kg: 3 students  

The frequencies are: 2, 1, 4, 2, 3.  

The highest frequency is 4 (at 58 kg).  Thus, Modal weight: 58 kg

Akọwa Nkọwa

To find the percentage profit, you first need to calculate two things:

• The total cost (how much the dealer spent altogether)
• The profit (how much more money he got from selling than he spent)

Step 1: Calculate the total cost.
The dealer bought the car for ₦270,000 and spent ₦70,000 to refurbish it.
So, the total cost is:

Step 2: Calculate the profit.
He sold the car for ₦490,000. Profit is the amount received from selling minus the total cost:

Step 3: Calculate the percentage profit.
Percentage profit is found by dividing the profit by the total cost and then multiplying by 100 to get a percentage:

Step 4: Work out the percentage:

Therefore, the percentage profit is approximately 45%.

Key Points:

• Percentage profit is always calculated on the total amount spent (cost price), not on the selling price.
• It's important to include both the purchase price and the refurbishment cost in the total cost before finding the profit.

Akọwa Nkọwa

To integrate the function \( y = 3x^2 + 2x - 5 \) with respect to \( x \), you need to find the indefinite integral:

\[ \int (3x^2 + 2x - 5)\,dx \]

We can integrate each term separately:

• \( \int 3x^2\,dx \): The integral of \( x^n \) is \( \frac{x^{n+1}}{n+1} \), so:
  \[ \int 3x^2\,dx = 3 \times \frac{x^{2+1}}{2+1} = 3 \times \frac{x^3}{3} = x^3 \]
• \( \int 2x\,dx \): Similarly,
  \[ \int 2x\,dx = 2 \times \frac{x^{1+1}}{1+1} = 2 \times \frac{x^2}{2} = x^2 \]
• \( \int -5\,dx \): The integral of a constant \( a \) with respect to \( x \) is \( a x \), so:
  \[ \int -5\,dx = -5x \]

Now, combine these results and add the constant of integration \( C \):

\[ x^3 + x^2 - 5x + C \]

This is the general antiderivative for the given function. Each term has been integrated correctly using the power rule of integration and the constant of integration \( C \) is included to represent all possible antiderivatives.

Akọwa Nkọwa

The question gives the value of \(\cos \theta\) as \(\frac{x}{y}\), and asks for \(\tan \theta\) in terms of \(x\) and \(y\).

Recall the definitions from trigonometry in a right triangle:

• \(\cos \theta = \frac{\text{adjacent side}}{\text{hypotenuse}}\)
• \(\sin \theta = \frac{\text{opposite side}}{\text{hypotenuse}}\)
• \(\tan \theta = \frac{\text{opposite side}}{\text{adjacent side}}\)

So, if \(\cos \theta = \frac{x}{y}\), then:

• The adjacent side has length \(x\).
• The hypotenuse has length \(y\).

We need the opposite side to find \(\tan \theta\). Use the Pythagorean theorem for a right triangle:

Plug in the values:

Solve for the opposite side:

Now, put this into the ratio for \(\tan \theta\):

This formula comes from the relationships among the sides in a right triangle. It uses the fact that the square of the hypotenuse minus the square of the adjacent side gives the square of the opposite side, and then takes the square root to get its length. That's why \(\tan \theta\) in terms of \(x\) and \(y\) is \(\frac{\sqrt{y^2 - x^2}}{x}\).