Supplementary angles are two angles whose measures add up to 180 degrees. e.g. 30, 150

2x² + 7x + 3

= 2x² + 6x + x + 3

= 2x (x+3) +1(x+3)

= (2x+1)(x+3)

{tex}5-\sqrt 6\over 5 + \sqrt 6{/tex}

Rationalize the denominator

= {tex}{5-\sqrt 6\over 5 + \sqrt 6}\times{ 5-\sqrt 6\over 5 -\sqrt 6}{/tex}

= {tex}25+6-10\sqrt 6\over 25-6{/tex}

= {tex}31 - 20\sqrt 6\over 19{/tex}

4x³ + 3x² - 4x + k

x-1 is factor of given polynomial

So x = 1ll be zero of this polynomial

So

p(1) = 0

=> 4(1)³ + 3(1)² - 4(1) + k = 0

=> 4 + 3 - 4 + k = 0

=> 3 + k = 0

=> k = -3

Four rational numbers between {tex}{{ - 2} \over 3}{/tex} and {tex}{{ 3} \over 2}{/tex}

Making both rational numbers as like terms,

{tex}{{ - 2} \over 3} \times {2 \over 2}{/tex} and {tex}{3 \over 2} \times {3 \over 3}{/tex}

{tex}\Rightarrow{/tex}         {tex}{{ - 4} \over 6}{/tex} and {tex}{9 \over 6}{/tex}

Now we can choose any four rational numbers between these rational numbers

{tex}{{ - 3} \over 6},{{ - 2} \over 6},{{ - 1} \over 6},{0 \over 6},{1 \over 6},{2 \over 6},{3 \over 6},{4 \over 6},.....,{8 \over 6}{/tex}

Ans. A = {tex}x+3\sqrt 3{/tex}

B = {tex}1\over A^2{/tex} = {tex}1\over {\left( {x+3 \sqrt 3}\right )^2}{/tex} 

Now {tex}A^2 = ({x+3 \sqrt 3})^2 \ ............... (1){/tex}

{tex}B^2 = ({1\over( x+3\sqrt 3)^2})^2 = {1\over (x+3\sqrt3)^4} \ ...... (2){/tex}

Adding (1) and (2), we get 

=> {tex}A^2 + B^2 = {(x+3\sqrt 3)^2} + {1\over (x+3\sqrt 3)^4} {/tex}

=> {tex}{(x+3\sqrt 3)^2(x+3\sqrt 3)^4 + 1 \over (x+3\sqrt 3)^4}{/tex}

{tex}=> {(x+3\sqrt 3)^6 +1 \over (x+3\sqrt 3)^4}{/tex}

Ans. Irrational number b/w {tex}\sqrt 2{/tex}  and 3

There are many numbers, you can choose any 

e.g {tex}\sqrt {2.1}, \sqrt {2.2} , \sqrt {3}{/tex}

Question should be Show that {tex}(a^{x+y})^{x-y} .(a^{y+z})^{y-z} .(a^{z+a})^{z-a} = 1{/tex}

Taking LHS

{tex}(a^{x+y})^{x-y} .(a^{y+z})^{y-z} .(a^{z+a})^{z-a}{/tex}

Using {tex}(x^m)^n= x^{mn}{/tex}

= {tex}a^{x^2-y^2} .a^{y^2-z^2} .a^{z^2-a^2}{/tex}

= {tex}a^{x^2-y^2+y^2-z^2+z^2-a^2}{/tex}

Using {tex}(x)^m.(x)^n= (x)^{m+n}{/tex}

= a⁰ = 1 = RHS

Verified 

In geometry, Heron's formula (sometimes called Hero's formula), named after Hero of Alexandria, gives the area of a triangle by requiring no arbitrary choice of side as base or vertex as origin.

Let a,b, and c are sides of a triangle. 

Then s = (a+b+c) ÷ 2

s is semi-perimeter of triangle

Then According to Heron formula

 area of triangle = {tex}\Delta = \sqrt {s(s-a)(s-b)(s-c)}{/tex}

let root 6 be rational no.

root 6 = a/b  where a and b are integers , b not equal to zero and also a and b are coprime

6 = a²/b²   (squaring both sides )

6b² = a²

b² = a²/6

therefore 6 is a factor of a ( 6 divides a² 

                                           so 6 divides a )

let us take a= 6c

now 6b²= a²

6b² = (6c)²    (putting value of a )

6b² = 36c²

c² = 6b²/36

c² = b²/6

therefore 6 is also a factor of b ( 6 divides b²

                                                 so 6 divides b )

so 6 is a factor of a as well as b .this contradicts the fact that a and b are coprime . this is due to our assumption

that root 6 is rational

                                                        THEREFORE  ROOT 6 IS IRRATIONAL

Solution.

Multiply and divide both number by 10,

we get

{tex}{10\over 70}, {20\over 70}{/tex}

Rational Numbers in between these two are{tex}{11\over 70}, {13\over 70}{/tex}

Ans. While it is true that any irrational number has a non-terminating decimal representation, the converse is not true.
For example,{tex}1\over 3{/tex} is a rational number whose decimal representation is non-terminating.

The correct statement is that a number whose decimal representation is non-terminating and non-repeating is irrational.

Refer Google step by step instructions are available there.

   ax² + bx + c 

= ax² + ex + fx + c ( where ex and fx are                                         factors of ac and their                                      sum is equal to bx ).

Example:

1.   x² + 7x + 10

= factors of 10 are 5 × 2. Also, 5 + 2 = 7

= hence, x² + 5x + 2x + 10

= x (x+5) + 2 (x+5)

= (x+5)(x+2)

2. 2x² + 12x + 10

= factors of 20 are 10 × 2. Also, 10+2 = 12.

= Hence, 2x² +10x + 2x + 10

= 2x (x + 5)+2 (x+5)

= (x+5)(2x+2)

Hope it helps!

The correct expression is

x²y - y²x - z²y + y²z + z²x - x²z

= x²y - x²z - y²x + z²x - z²y + y²z

= x²(y - z) - x(y² - z²) + yz(y - z)

= x²(y - z) - x(y - z)(y + z) + yz(y - z)

= (y - z){x² - xy - xz + yz}

= (y - z){x(x - y) - z(x - y)}

= (y - z)(x - y)(x - z)

= (x - y)(y - z)(x - z)

4/9 and 7/11

LCM of 9 and 11 is 99.

Hence, 

     4 × 11/ 9 × 11 = 44/99

     7 × 9/ 11 × 9 = 63/99

 2 rational numbers between 4/9 and 7/11 are 45/99 and 46/99

(1)

Dimensions of room :

l = 5m ; b = 3.5m  ; h= 3m

LSA of room = 2(l+b) x h

= 2 ( 5 + 3.5) x3

= 6 x 8.5

= 52 m²

1 m² = Rs. 20

52 m² = Rs. 20 x 52 = Rs. 1040

(2) 

Let the edge of the cube be 'a' 

Surface area = 6a²

Let the edge be increased by 50% so the new edge becomes = a + (a/2)  = 3a/2

New Surface area = 6 (3a/2)² = 27a²/2

Percentage increase = [Change in Increase/ Original] x 100

= [(27a² - 12a²)/ 6a² x 2 ] x 100

(5 x 100 )/4

5 x 25  = 125%

{tex}{4\over (216)^{-2\over 3}} - {1\over (256)^{-3\over 4}}{/tex}

{tex}= {4\over( (6)^3)^{-2\over 3}} - {1\over ((4)^4)^{-3\over 4}}{/tex}

{tex}= {4\over (6)^{-2}} - {1\over (4)^{-3}}{/tex}

{tex}= 4\times 36 - 64 = 144-64 = 80{/tex}

k²x³ - kx² + 3k - k = 0

=>  k²(3)³ - k(3)² + 3k - k = 0

=>  27k² - 9k + 3k - k = 0

=>  27k² - 7k = 0

=>  k(27k - 7) = 0

=> k = 0 or 27k - 7 = 0

=> k = 0 or k = 7/27

0.51 , 0.52 , 0.53 , 0.54 , 0.55

Since,

{tex}{x^3} + {y^3} + {z^3} - 3xyz = \left( {x + y + z} \right)\left( {{x^2} + {y^2} + {z^2} - xy - yz - zx} \right){/tex}

Putting x + y + z = 0 [Given]

=>          {tex}{x^3} + {y^3} + {z^3} - 3xyz = \left( 0 \right)\left( {{x^2} + {y^2} + {z^2} - xy - yz - zx} \right){/tex}

=>          {tex}{x^3} + {y^3} + {z^3} - 3xyz = 0{/tex}

=>          {tex}{x^3} + {y^3} + {z^3} = 3xyz{/tex}

{tex}0x+3y+5=0{/tex}

Mean = Sum of observations / Total number of observations

=>          24 = Sum of observations / 100

=>          Sum of observations = 24 x 100 = 2400

Now new sum = 2400 + 100 x 6 = 2400 + 600 = 3000

Also, New sum = 3000 x 2.5 = 7500

Then, New Mean = 7500/100 = 75

mean of 100 observations = 26

mean after adding 6 to each observation = 24+6 = 30

New mean after multiplying 2.5 = 30 x 2.5 = 75.

Total surface area of cube = 96 cm²

6a² = 96 cm²

a² = 96/6 = 16 cm²

a = 4 cm

volume of cube = a³

⁼ 4³ = 64 cm³

By prime factorization or long division method

No.because it is nor terminating and not repeating number