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Let the speed of X and Y be x km/hr and y km/hr respectively.

We know that the time taken to cover 'd' km with speed 's' km/hr is {tex}\frac ds{/tex} hr.

Now, using above formula for time 
Time taken by X to cover 30 km {tex}= \frac { 30 } { x } \mathrm { hrs }{/tex}
and, Time taken by Y to cover 30 km {tex}= \frac { 30 } { y } \mathrm { hrs }{/tex}
By the given condition in 1st half of problem, we have 
{tex}\frac { 30 } { x } - \frac { 30 } { y } = 3 \Rightarrow \frac { 10 } { x } - \frac { 10 } { y } = 1{/tex} ................................(i)
If X doubles his pace, then speed of X becomes 2x km/hr 
{tex}{/tex}Time taken by X to cover 30 km {tex}= \frac { 30 } { 2 x } \mathrm { hrs }{/tex}
&, Time taken by Y to cover 30 km {tex}= \frac { 30 } { y } \mathrm { hrs }{/tex} (speed of Y remains constant)
According to the given condition in 2nd half of problem, we have
{tex}\frac { 30 } { y } - \frac { 30 } { 2 x } = 1 \frac { 1 } { 2 }{/tex}
{tex}\Rightarrow \quad \frac { 30 } { y } - \frac { 30 } { 2 x } = \frac { 3 } { 2 }{/tex}
{tex}\Rightarrow \quad \frac { 10 } { y } - \frac { 5 } { x } = \frac { 1 } { 2 }{/tex}
{tex}\Rightarrow \quad - \frac { 10 } { x } + \frac { 20 } { y } = 1{/tex}.................................(ii)
Putting {tex}\frac { 1 } { x } = u{/tex} and {tex}\frac { 1 } { y } = v,{/tex}in equations (i) and (ii) we get
10{tex}u{/tex} -10{tex}v{/tex} =1 {tex}\Rightarrow{/tex}10{tex}u{/tex}- 10{tex}v{/tex}-1=0 ........(iii)
-10{tex}u{/tex}+ 20{tex}v{/tex} = 1 {tex}\Rightarrow{/tex} -10{tex}u{/tex} + 20{tex}v{/tex} -1=0 ........(iv)
Adding equations (iii) and (iv), we get
{tex}10 v - 2 = 0 \Rightarrow v = \frac { 1 } { 5 }{/tex}
Putting {tex}v = \frac { 1 } { 5 }{/tex} in equation (iii), we get
{tex}10 u - 3 = 0 \Rightarrow u = \frac { 3 } { 10 }{/tex}
Now, {tex}u = \frac { 3 } { 10 } \Rightarrow \frac { 1 } { x } = \frac { 3 } { 10 } \Rightarrow x = \frac { 10 } { 3 }{/tex}and {tex}v = \frac { 1 } { 5 } \Rightarrow \frac { 1 } { y } = \frac { 1 } { 5 } \Rightarrow y = 5{/tex}
Hence, X's speed {tex}= \frac { 10 } { 3 } \mathrm { km } / \mathrm { hr }{/tex} and, Y's speed =5km/hr.

Assume f(x) = {tex}6{x^3} + \sqrt 2 {x^2} - 10x - 4\sqrt 2{/tex}
If {tex}\sqrt 2{/tex} is the zero of f(x), then {tex}(x - \sqrt 2 ){/tex} will be a factor of f(x). So, by remainder theorem when f(x) is divided by {tex}(x - \sqrt 2 ){/tex}, the quotient comes out to be quadratic.
Now we divide  {tex}6{x^3} + \sqrt 2 {x^2} - 10x - 4\sqrt 2{/tex} by {tex}(x - \sqrt 2 ){/tex}.

{tex}\therefore \;f(x) = (x - \sqrt 2 )(6{x^2} + 7\sqrt 2 x + 4){/tex} (By Euclid’s division algorithm)
{tex}= (x - \sqrt 2 )(6{x^2} + 4\sqrt 2 x + 3\sqrt 2 x + 4){/tex} ( By factorization method )
For zeroes of f(x), put f(x) = 0
{tex}\therefore (x - \sqrt 2 )(6{x^2} + 4\sqrt 2 x + 3\sqrt 2 x + 4) = 0{/tex}
{tex}\Rightarrow \;(x - \sqrt 2 )[2x(3x + 2\sqrt 2 ){/tex} {tex}+ \sqrt 2 (3x + 2\sqrt 2 )] = 0{/tex}
{tex}\Rightarrow (x - \sqrt 2 )(3x + 2\sqrt 2 )(2x + \sqrt 2 ) = 0{/tex}
{tex}\Rightarrow x - \sqrt 2 = 0{/tex} or {tex}3x + 2\sqrt 2 = 0{/tex} or {tex}2x + \sqrt 2 = 0{/tex}
{tex}\Rightarrow x = \sqrt 2{/tex} or {tex}x = \frac{{ - 2\sqrt 2 }}{3}{/tex} or {tex}x = \frac { - \sqrt { 2 } } { 2 }{/tex}
So, other two roots are {tex}= \frac{{ - 2\sqrt 2 }}{3}{/tex} and {tex}\frac{{ - \sqrt 2 }}{2}{/tex}.

Sin60° + cos30°
(√3/2)  + (√3/2)
= ( (√3+ √3)/ 2
= 2√3/2
= √3

Given integers,
595 and 252
Applying Euclid division algorithm to 595 and 252 we get,
595=252×2+91.........................................1

Now applying ,Euclid division algorithm to 252 and 91 we get,
252=91×2+70.....….………...........................2

Now applying Euclid division algorithm to 91 and 70,we get
91=70×1+21…...............................................3

Now applying Euclid division algorithm to 70 and 21 we get,
70=21×3+7...................................................4

Now applying Euclid division algorithm to 21 and 7 we get,
21=7×3+0......................................................5
The remainder at this stage is zero.
Hence HCF of 595 and 252 is 7.

eq1,eq2,eq3, and eq4 can be written as,
595-(252×2)=91..........................................6

252-(91×2)=70.............................................7

91-(70×1)=21...................................................8

70-(21×3)=7.....................................................9
Now eq9 we have,
(as we express HCF in the form of a equation that is the reason we start from eq9)
7=70-(21×3)
By putting eq8 in above equation we have,
7=70-(91-70×1)3

7=70-91×3+70×3

7=70×4-91×3
By putting eq7 in above equation we have,
7=(252-91×2)4-91×3

7=252×4-91×8-91×3

7=252×4-91×11
By putting eq6 in above equation we have,
7=252×4-(595-252×2)11

7=252×4-595×11+252×22

7=252×26-595×11

7=595(-11)+252(26)

7=595m+252n, where m=-11 and n=26.
Hence HCF of 595 and 252 is in the form of 595m+252n,where m=-11 and n=26.

Since the circular ends of the diameter are 28 cm and 42 cm, the radii of are 14 cm and 21 cm respectively.
Capacity of the bucket {tex}= \frac { 1 } { 3 } \pi h\left( \mathrm { R } ^ { 2 } + \mathrm { r } ^ { 2 } + \mathrm { Rr } \right){/tex}
{tex}= \frac { 1 } { 3 } \times \frac { 22 } { 7 } \times 24 \left[ 14 ^ { 2 } + 21 ^ { 2 } + 14 \times 21 \right]{/tex}
{tex}= \frac { 22 } { 7 } \times 8 [ 196 + 441 + 294 ]{/tex}
{tex}= \frac { 22 } { 7 } \times 8 \times [ 196 + 441 + 294 ]{/tex}
= 23408 cm³
= 23.408 l ...(Since 1000 cm³ = 1 l)
Cost of milk at the rate of Rs.30 per litre
= 23.408{tex}\times{/tex}30
= Rs. 702.24

Any integer can be written in the form 5m , 5m+1, 5m+2 (where m is any integer)

(5m)² = 25m² = 5 × 5m² = 5q (where q = 5m²)

(5m+1)² = (5m)² + 2 ×5m × 1 + 1² [by using identity- (a+b)²= (a²+ 2ab +b² )]
       = 25m² + 10 m + 1
        = 5 (5m²+ 2m) +1
        = 5q +1 ( where q= 5m²+2m)
(5m+2)²= (5m)² + 2× 5m × 2 +2² [by using identity- (a+b)²= (a²+ 2ab +b² )]
       = 25m² + 20 m +4
       = 5 (5m²+4m ) + 4
        = 5q+4 (where q= 5m²+4m)
Hence proved

If a and b are odd numbers then it should be in 2q+1 or 2q+3 form where q is a positive integer.

Let a = 2q + 3 , b = 2q + 1  and a > b
Now, {tex}\frac{a + b } {2} = \frac{ 2q + 3 + 2q + 1}{2}{/tex}
{tex}= \frac { 4 q + 4 } { 2 }{/tex}
= 2q + 2
{tex}\frac{a+b}{2}{/tex}=2(q+1) = an even number..........(1)
Now

 {tex}\frac { a - b } { 2 } = \frac { ( 2 q + 3 ) - ( 2 q + 1 ) } { 2 }{/tex}
{tex}= \frac { 2 q + 3 - 2 q - 1 } { 2 }{/tex}
{tex}\frac{a-b}{2}= \frac { 2 } { 2 }{/tex} = 1 = an odd number..........(2)
Hence From (1) and (2) {tex}\frac{a+b}{2}{/tex}   and {tex}\frac{a-b}{2}{/tex}   are even and odd numbers respectively

Cramer's Rule is a method that uses determinants for solving systems of linear equations. To explain the method and how these determinants are generated, this lecture will first illustrate the solution to a system of linear equations by the addition/elimination method without simplifying the indicated products and sums of the coefficients. The intent is to demonstrate that these products and sums can be represented by determinants.

A polynomial is a mathematical expression that consists of variables and constants combined using addition, subtraction and multiplication. Variables may have non-negative integer exponents. Although division by a constant is allowed, division by a variable is not allowed.

The degree of a polynomial is the exponent of the highest degree term.

• A polynomial of degree 0 is called a constant polynomial.
• A polynomial of degree 1 is called a linear polynomial.
• A polynomial of degree 2 is called a quadratic polynomial.
• A polynomial of degree 3 is called a cubic polynomial.