Since 11 is not a perfect square \(\sqrt{11}\)

Cannot be written as a rational number and is therefore irrational. In contrast, \(\sqrt{16}=4, \sqrt{49}=7 .\)and

\(\sqrt{100}=10\) are rational because 16, 49, and 100 are perfect squares

The midpoint between \(\frac{1}{4} \text { and } \frac{1}{3}\) is found by taking their average:

\(\begin{aligned} & \text { Midpoint }=\frac{\frac{1}{4}+\frac{1}{3}}{2} \\ & \frac{1}{4}+\frac{1}{3}=\frac{3}{12}+\frac{4}{12}=\frac{7}{12} \\ & \frac{7}{12} \div 2=\frac{7}{12} \times \frac{1}{2}=\frac{7}{24} \end{aligned}\)

So, the midpoint is \(\frac{7}{24}\)

\(\frac{7}{0}\) is undefined because division by zero is not possible in mathematics. In any rational number \(\frac{p}{q},\) the denominator must satisfy \(q \neq 0 .\)

If \(\frac{7}{0}=x, \text { then } x \times 0=7\)

But any number multiplied by zero equals zero, never 7. Therefore, no such number x exists. Hence, division by zero is undefined not a rational number.

Brahmagupta explained negative numbers using everyday trade language. He called positive numbers Dhana (wealth) and negative numbers Rina (debts). For example, if you have ₹ 3 but owe ₹ 5, you are ₹ 2 in debt i.e., 3 -5=-2.

The sum of consecutive interior angles on the same side of a transversal is \(180^{ 0 }\)

\(x+50^{ 0 }\) 
\(y+90^{ 0 }+x=180^{ 0 }\quad 50^{ 0 }+90^{ 0 }+x=180^{ 0 }\) 
\(\Rightarrow x=40^{ 0 }\) 
\(q=x=40^{ 0 }\) 

\(4x+2x=180^{ 0 }\)

\((30-a)^{ 0 }+(125+2a)^{ 0 }=180^{ 0 }\)

\(150^{ 0 },30^{ 0 }=180^{ 0 }\neq 90^{ 0 }\)

Let AB and CD two lines intersecting at O

This leads to two pairs of vertically  opposite angles,namely
(i)  \(\angle \)AOC and \(\angle \)BOD
(II) \(\angle \)AOD and \(\angle \)BOC
We are to prove that
(i)   \(\angle \)AOC and \(\angle \)BOD
(II) \(\angle \)AOD and \(\angle \)BOC
\(\therefore \) Ray OA stands on line CD 
Therefore (i)   \(\angle \)AOC and \(\angle \)AOD=\(180^{ 0 }\)
|Linear Pair Axiom
From (1) and (2) 
\(\angle \)AOC and \(\angle \)AOD= \(\angle \)AOD and \(\angle \)BOD \(\angle \)AOC and \(\angle \)BOD
\(\Rightarrow \) Similarly we can prove that 
\(\angle \)AOD and \(\angle \)BOC

30°, 60°, 90°.

Since ray OP stands on line AB.
∴\(\angle\)POA + \(\angle\)POB = 180° [by linear pair axiom]
\(\Rightarrow\) \(\angle\)POA + [\(\angle\)POQ + \(\angle\)QOR + \(\angle\)ROB] = 180°
[\(\because\) \(\angle\)POB = \(\angle\)POQ + \(\angle\)QOR + \(\angle\)ROB]
Hence, the sum of all the angles on the same side of a straight line at a given point is 180°.
Hence proved.

\(255^{ 0 }\)

30

50

∵ AB ॥ CD and EF is a transversal.
∴ ∠BEP = ∠EFD [corresponding angles]
⇒ \(\frac{1}{2}\) ∠BEP = \(\frac{1}{2}\) ∠EFD
⇒ ∠PEG = ∠EFH [∵ EG and FH are the angle bisectors of ∠BEP and ∠EFD respectively]
But they form pair of corresponding angles.
∴ EG॥ FH.

5x-20^o+2x-10^o=180^o (Corresponding interior angles)
\(\Rightarrow\)7x=180^o+30^o=210^o
\(\Rightarrow\)x=30^o
y=180^o-(5x-20^o)
=180^o-(150^o-20^o)
\(\Rightarrow\)y=180^o-130^o=50^o
z=2x-10^o
=60^o-10^o=50^o

Let us assume \(5+2 \sqrt{3}\) is rational, then it must be in the form of

where p and q are co-prime integers and \(q \neq 0\)

i.e., \(5+2 \sqrt{3}=\frac{p}{q}\)

So

Since p,g, 5 and 2 are integers and \(q \neq 0,\)

RHS of equation (i) is rational. But LHS of (i) is \(\sqrt{3}\) which is irrational.

This is not possible. This contradiction has arisen due to our wrong assumption that \(5+2 \sqrt{3}\) is rational.

So, \(5+2 \sqrt{3}\) is irrational

(A) x=3.0525252.., 10x =30.5252.

1000 x 3052.5252

990 x = 3022;

\(x=\frac{3022}{990}=\frac{1511}{495}\)

(B) X= 0.2323..

100x = 23.2323...

\(99 x=23 ; x=\frac{23}{99}\)

(C) X= 2.0555...

10x = 20.555...

100x =205.555...

\(90 x=185 ; x=\frac{185}{90}=\frac{37}{18}\)

(abscissa of P) - (abscissa of Q)

= (-2) - (-3) =-2+3=1

Let the circle meet OX at A and OX' at A'.

Then,

A→(4,0)

A'(-4,0)

Let the circle meet OY at B and OY' at B',

Then,

B→(0,4)

B'→ (0,-4)

(a) Food court → (-8,6)

(b) Space ride → (6, 2)

(c) Park entrance → (-10,-8)

(d) Water ride → (2, -2)

(i) (99)³ = (100 – 1)³ = (100)³ - (1)³ - 3(100)(1)(100 – 1) I Using Identity VII

= 1000 000 - 1 – 300(100 - 1) = 1000 000 - 1 - 30 000 + 300 = 970 299

(ii) (102)³ = (100 + 2)³ = (100)³ + (2) + 3(100)(2)(100 + 2) I Using Identity VI

= 1000000 + 8 + 600(100 + 2) = 1000000 + 8+ 60000 + 1200 = 1061208

(iii) (998)³ = (1000 - 2)³

= (1000)³ - (2)³ - 3(1000)(2)(1000 - 2) I Using Identity VII

= 100 00 00 000 - 8 - 6000(1000 - 2)

= 100 00 00 000 - 8 - 60 00 000 + 12000 = 99 40 11 992.

(A) \(\sqrt{169}=13\) ⇒ Rational (169 = 13², perfect square)

(B) \(\sqrt{75}=5 \sqrt{3}\) ⇒ Irrational (75 not a perfect square)

(C) 0.12112111211112... ⇒ Irrational (pattern grows no fixed block repeats)

(D) 3.205 ⇒ Rational (non-terminating recurring repeating block '05')

Since.

(A) Representing \(\frac{-7}{4}\) on the number line

Converting to a mixed number:

\(-\frac{7}{4}=-1 \frac{3}{4}\)

Since \(\frac{-7}{4}=-1.75 \text {, it lies between }-2 \text { and }-1 \text {. }\)

Construction steps:

(1) Draw a number line and mark the integers -2 and-1 on it.

(2) Divide the segment between -2 and -1 into 4 equal parts (because the denominator is 4).

(3) Each small part represents \(\frac{1}{4} .\)

(4) Starting from -2, move 1 part to the right (since

\(\left.-2=\frac{-8}{4} \text { and } \frac{-7}{4}=\frac{-8}{4}+\frac{1}{4}\right)\)

(5) The 1st division point to the right of -2 represents \(\frac{-7}{4} .\)

(B) Representing \(\frac{5}{6}\) on the number line

Checking the value:

\(\frac{5}{6} \approx 0.833\)

Since \(0<\frac{5}{6}<1 \text {, it lies between } 0 \text { and } 1 .\)

Construction steps:

(1) Draw a number line and mark the integers 0 and 1 on it

(2)Divide the segment between 0 and 1 into 6 equal parts (because the denominator is 6).

(3) Each small part represents

(4) Starting from 0, move 5 equal parts to the right.
(5) The 5^th division point to the right of O represents \(\frac{5}{6}\)

Given fractions: \(\frac{2}{3} \text { and } \frac{4}{6}\)

Check cross multiplication:

2 x 6 = 12, 3 x 4 = 12

Since ad= bc, therefore:

Hence \(\frac{2}{3}=\frac{4}{6}\)

Also,

\(\begin{aligned} \frac{2}{5}+\left(-\frac{3}{10}\right) & =\frac{2}{5} \times\left(-\frac{10}{3}\right)=-\frac{20}{15}=-\frac{4}{3} \\ & =-\frac{20}{15}=-\frac{4}{3} \end{aligned}\)

(A) (-12) x 5 = -60 ⇒ debt x fortune = debt

(B) (-8) x (-7) =56 debt x debt=fortune

(C) 0 - (-14) = 0 + 14 = 14 ⇒ subtracting a debt = adding a fortune

(D) (-20) ÷ 4 = -5 debt ÷ fortune = debt

The coordinates of the point are (x,y).

The general form of any point on the x-axis is (x,0).

(i) (-12)³ + (7)³ + (5)³ = 3(- 12)(7)(5) =-1260 |∵ (-12) + (7) + (5) = 0

(ii) (28)³ + (-15)³ + (-13)³ = 3(28)(- 15)(- 13) = 16380. | ∵ (28) + (- 15) + (- 13) = 0

Both the cross-streets are marked in the figure given. They are uniquely found because of the two reference lines we have used for locating them.

(i) 9x² + 6xy + y² = (3x)² + 2(3x)(y) + (y)² = (3x+ y)² = (3x + y)(3x + y) I Using Identity I

(ii) 4y² - 4y + 1 = (2y)² - 2(2y)(1) + (1)² = (2y - 1)² = (2y - 1)(2y - 1) I Using Identity II

(iii) \(x^2-\frac{y^2}{100}\)= \((x)^2-\left(\frac{y}{10}\right)^2=\left(x+\frac{y}{10}\right)\left(x-\frac{y}{10}\right) .\) | Using Identity III

(a) Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

Explanation:

Assertion (A) is true because an irrational number always exists between any two rational numbers. For example \(\sqrt{2}\) lies between 1 and 2 and is irrational. Reason (R) is also true because rational numbers are dense but do not cover the entire number line; irrational numbers occupy the remaining points. This explains why an irrational number can always be found between any two rational numbers.

(c) Assertion is correct but Reason is incorrect.

Explanation:

Hence, the product is rational. However, the reason is incorrect because the product of two irrational numbers is not always irrational. This example itself is a counter example

(A) Prove that \(\sqrt{10}\) is irrational.

Assume that \(\sqrt{10}=\frac{p}{q}\)

where p and q are co-prime integers and q ≠ 0.

Squaring both sides, p² =10q²

Since 10 = 2 x 5, both 2 and 5 divide p².
Therefore, 2 divides p and 5 also divides p. Hence, 10 divides p.

Let p = 10m

Substituting in the equation:

(10m)² = 10q²

100m² = 10q²

q² = 10m²

Thus, 10 also divides q.

So both p and q are divisible by 10, which contradicts the assumption that p and q are co-prime.

Hence, \(\sqrt{10}\) is irrational.

(B) For \(\sqrt{4}: p^2=4 q^2\)

Taking p = 2k,

(2k)² = 4q²

4K² = 4q²

k = q

Therefore,

\(\sqrt{4}\) = 2 which is rational,

Similarly, for \(\sqrt{9}:\)

p² = 9q²

Taking p = 3k,

9k² = 9q²

So k = q, giving

\(\sqrt{9}=3\)

which is also rational.

The proof by contradiction works only when the number is not a perfect square. For perfect squares such as and 9, the steps simplify to a rational number instead of producing a contradiction.