The distance between two points \( (x_1, y_1)\) and \( (x_2, y_2)\) is given by the formula:

\( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)

Substitute the values of the given points \((x_1, y_1) = (2, -1)\) and \((x_2, y_2) = (-1, -5):\)

\( d = \sqrt{((-1) - 2)^2 + ((-5) - (-1))^2} \)

\( d = \sqrt{(-3)^2 + (-4)^2} \)

\( d = \sqrt{9 + 16} \)

\( d = \sqrt{25} = 5 \)

\(\therefore\) The distance between the points is 5 units.

The formula for the midpoint \( \text {M}(x_m, y_m) \) of the line segment joining two points \(\text { A}(x_1, y_1) \) and \( \text {B}(x_2, y_2) \) is:

\( x_m = \dfrac{x_1 + x_2}{2}\), \(\quad y_m = \dfrac{y_1 + y_2}{2} \)

Given that \( \text {C}(1, -1) \) is the midpoint of the line segment joining points \( \text {A}(4, x) \) and \(\text { B}(-2, 4) \), we use the midpoint formula.

For the \(x\)-coordinate:

\( 1 = \dfrac{4 + (-2)}{2}\) \(= \dfrac{4 - 2}{2} = \dfrac{2}{2} \) \(= 1 \)

Thus, the \(x\)-coordinate is correct.

For the \(y\)-coordinate:

\( -1 = \dfrac{x + 4}{2} \)

Multiplying both sides by 2:

\( -2 = x + 4 \)

Solving for \(x\):

\( x = -2 - 4 = -6 \)

Therefore, the value of \(x\) is \(-6\).

In probability, the sum of the probability of an event \( {\text {E}}\) and its complement \(\overline {\text {E}}\) is always 1. This is because:

\(\text {P(E)}+ \text {P}(\overline {\text {E}}) = 1\)

Option (C) is correct because it reflects this relationship.

1. Identify the modal class. The modal class is the class interval with the highest frequency. Here, the highest frequency is 14, corresponding to the class interval 425 − 449.

2. The lower limit of this modal class is the lower boundary of the class interval. Hence, the lower limit is:

Lower Limit = 424.5

1. Represent the problem using a right triangle, where the height of the lamp post is the opposite side, and the length of the shadow is the adjacent side.

2. Use the tangent of the angle of elevation: \( \tan(\theta) = \dfrac{\text{opposite}}{\text{adjacent}} \)

3. Substituting the values: \( \tan(\theta) = \dfrac{9}{3\sqrt{3}} \)

4. Simplify the fraction: \( \tan(\theta) = \dfrac{9}{3\sqrt{3}} = \dfrac{3}{\sqrt{3}} = \sqrt{3} \)

5. The angle \(\theta\) whose tangent is \(\sqrt{3}\) is \(60^\circ\).

6. Therefore, the Sun’s elevation is: \( \theta = 60^\circ \).

Let the given polynomial be \( {\text {k}x^2 + 4x +\text { k}}\). We know that one of the roots is 1.

So, substitute \(x = 1\) into the polynomial and set it equal to zero:

\( k(1)^2 + 4(1) + k = 0 \)

\( k + 4 + k = 0 \)

\( 2k + 4 = 0 \)

\( 2k = -4 \)

\( k = -2 \)

Thus, the value of \((\text {k})\) is \(-2\).

1. A quadratic polynomial with roots \(-1\) and \(3\) can be expressed as: \( k(x + 1)(x - 3), \) where \(k \neq 0\).

Here, \(k\) is any non-zero constant.

2. Since \(k\) can take infinitely many non-zero values, the number of quadratic polynomials with these zeroes is more than 3.

The given quadratic equation is

\( x^2 - 4 = 0, \)

which is a difference of squares:

\( x^2 - 4 = (x - 2)(x + 2) = 0 \)

Thus, the roots are

\( x = 2 \quad \text{and} \quad x = -2. \)

1. Expand each equation to check its degree:-

For (a), expanding gives a degree of 2, so it is a quadratic equation.-

For (b), expanding gives a degree of 2, so it is a quadratic equation.-

For (c), after expansion, the terms cancel out, and the degree becomes less than 2. Hence, it is not a quadratic equation.-

For (d), it contains \(\dfrac{1}{x}\) , making it not a polynomial, but not a quadratic equation either.

2. Thus, the correct answer is (c).

In an arithmetic progression (A.P.), the \(n\)-th term is given by the formula: \( a_n = a_1 + (n - 1)d \)

Given that \(a_{23} - a_{19} = 32\), we can use the formula for the \(n\)-th term: \( a_{23} = a_1 + 22d, \quad a_{19} = a_1 + 18d \)

Subtracting the two equations:

\( a_{23} - a_{19} = 32 \)

\( (a_1 + 22d) - (a_1 + 18d) = 32 \)

\( 22d - 18d = 32 \)

\( 4d = 32 \Rightarrow d = 8 \)

Thus, the common difference is 8.

1. Rewrite \(\tan^2 \theta\) in terms of \(\sin\) and \(\cos\): \( \tan^2 \theta = \dfrac{\sin^2 \theta}{\cos^2 \theta} = \dfrac{\cos^2 \theta}{\sin^2 \theta} \)

2. Subtract \(\sin^2 \theta\): \( \dfrac{\cos^2 \theta}{\sin^2 \theta} - 1 = \dfrac{\cos^2 \theta - 1}{\sin^2 \theta} \)

3. Using the identity \(\cos^2 \theta - 1 = -\sin^2 \theta\), substitute: \( \dfrac{\cos^2 \theta - 1}{\sin^2 \theta} = \dfrac{-\sin^2 \theta}{\sin^2 \theta} = -1 \)

4. Therefore, the value is \(-1\).

In a circle, the region between a chord and either of the two arcs is called a segment. A sector is the region between two radii and the corresponding arc, while an arc is simply a part of the circumference.

We first find the prime factorization of 1080: \( 1080 = 2^3 \times 3^3 \times 5 \)

Comparing this with the given equation \(1080 ={2^x} \times 3^y \times 5\), we get: \( x = 3, \quad y = 3 \)

Thus, \( x - y = 3 - 3 = 0. \)

We know that the radius of a circle is perpendicular to the tangent at the point of contact. Therefore:

\( \angle \text {OAP} = 90^\circ \)

We are given that:

\( \angle \text {AOP} = 70^\circ \)

Since \(\text {PA}\) is a tangent, the angles \(\angle \text {OAP}\) and \(\angle \text {APO}\) form a linear pair at point \(\text {P}\). Hence:

\( \angle \text {OAP} + \angle \text {APO }= 180^\circ \)

Substituting the known values:

\( 90^\circ + \angle \text {APO }= 180^\circ \)

\( \angle \text {APO} = 180^\circ - 90^\circ - 70^\circ = 20^\circ \)

Thus, the correct answer is: \( \text{(D) } 20^\circ \)

The median group corresponds to the group where the cumulative frequency exceeds half the total frequency. First, find the cumulative frequencies:

Cumulative Frequency: 5, 13, 33, 48, 55, 60

First, calculate the cumulative frequency:

Cumulative Frequency: 5, 5+8 = 13, 13+20 = 33, 33+15 = 48, 48+7 = 55, 55+5 = 60

The total frequency is 60, and half of this is 30. The cumulative frequency just greater than 30 is 33, which corresponds to the group 20 – 30.

Thus, the median group is: (B) 20 – 30

The length of an arc \(\text {L}\) is given by the formula:

\( L = \dfrac{\theta}{360^\circ} \times 2 \pi r \)

Where: \(\theta = 60^\circ\) (central angle), \(r = 21 \text{ cm}\) (radius)

Substituting the values:

\( L = \dfrac{60^\circ}{360^\circ} \times 2 \times \pi \times 21 \)

\( L = \dfrac{1}{6} \times 42 \pi \)

\( L = 7 \pi \approx 22 \text{ cm} \)

Thus, the length of the arc is: \( {22 \text{ cm}} \)

By definition, a tangent to a circle is a straight line that touches the circle at exactly one point. This point is known as the point of tangency.

Thus, the correct answer is: (A) one point only

Since the triangles are similar, we use the property that the corresponding sides of similar triangles are proportional. Therefore, we have :

Substitute the known values :

\(\dfrac{\text {AC}}{\text {QR}} = \dfrac{\text {BC}}{\text {RP}}\)

\(\dfrac{6}{3} = \dfrac{4.6}{x}\)

Now, solve for x :

\( 2 = \dfrac{4.6}{x}\)

\(x = \dfrac{4.6}{2} = 2.3 \text{ cm}\)

Thus, the value of x is 2.3 cm.

We are given two linear equations:

\(5x + 2y + 6 = 0\) (Equation 1)

\(7x + 9y = 18\) (Equation 2)

To find if these equations have infinitely many solutions, we check the condition for infinitely many solutions, which is:

\(\dfrac {a_1} {a_2} = \dfrac {b_1} {b_2} = \dfrac{c_1} {c_2}\)

For the given equations, we have:

\(\dfrac{5}{7}\), \( \dfrac{2}{9}\), \(\dfrac{6}{18}\)

\(\dfrac{5}{7} \neq \dfrac{2}{9}\) and \(\dfrac{6}{18} = \dfrac{1}{3}\)

Since the ratios do not match, these equations do not have infinitely many solutions.

Therefore, the assertion is incorrect.

The total number of faces on a die is 6, numbered 1 to 6. Since 8 is not one of the numbers on the die, the probability of getting 8 is 0.

Hence, the assertion is correct. Additionally, the reason states that the probability of an impossible event is zero, which is also correct.