{tex}4\pi {r^2} = 98.56{/tex}

=>     {tex}{r^2} = {{98.56 \times 7} \over {4 \times 22}}{/tex}

=>     {tex}{r^2} = 7.84{/tex}

=>     {tex}r=2.8{/tex} cm

Volume of this box={tex} 4cm×2.5cm×1.5cm {/tex}

                             = {tex}15 cm^3{/tex}

Volume of 12 such boxes be {tex}15 × 12 cm^3{/tex} ={tex}180 cm^3{/tex}

x² + (4 + 10)x + (4)(10)

= x^{<font size="2">2</font>} + 14x + 40

sqrt of 53

53 is not a perfect square and contains no perfect squares,

so sqrt 53 = 7.280 (approximately).

area of parallelogram =Base ×Altitude =CD×AD=16×8=128cm^2.  (AB=CD )

Area of parallelogram=AD×CF=AD×10

Area of parallelogram= AD×10

128=10AD. (Area of parallelogram remain same either change of base and altitude )

AD=12.8cm

{tex}(2{\sqrt 3}+5{\sqrt 5}-7{\sqrt 7})+(3{\sqrt 5}-{\sqrt 3}+{\sqrt 7}){/tex}

={tex}2{\sqrt 3}-{\sqrt 3}+5{\sqrt 5}+3{\sqrt 5}-7{\sqrt 7}+{\sqrt 7}{/tex}

={tex}{\sqrt 3}(2-1)+{\sqrt 5}(5+3)+{\sqrt 7}(-7+1){/tex}

={tex}{\sqrt 3}×1+{\sqrt 5}×8+{\sqrt 7}×(-6){/tex}

={tex}{\sqrt 3}+8{\sqrt 5}-6{\sqrt 7}{/tex}

L. H. S  =   a^m / aⁿ  . b ⁿ / b^l  . C ^l / c ^m

               [ x ^m+n y^l  ] ^m  . [ x ^n+l y ^m ] ⁿ . [ x ^{l+ m} yⁿ ]^l

           =------------------------------------------------

               [ x ^m+n  y^l  ]ⁿ ​​​​​​. [ x ^n+l . y ^m ] ^l  [ x ^{l + m}  yⁿ ]^l 

^{multiply  m+n with  m,  l with m,  n+ l with n,  m with n} 

^{then split  numerator and denominator} 

^{all terms will cancel and remain 1}

^{.....thank you} 

^{all the best} 

So if polynomial p(x)= 4+x²-x+2x³

Then at x=-2

p(-2)=4+(-2)²-(-2)+2(-2)³

        =4+4+2+2×(-8)

        =10-16

        =-6

Let p(x)=4+x²-x+2x²

                =3x²-x+4

At x=-2

p(-2)= 3×(-2)²-(-2)+4

        =3×4+2+4

        =12+2+4

        =18

Since we need to find 5 rational nubers between 1 and 2 so we will multiply and divide both numbers by 5+1 =6 

So 1=1×{tex}{6\over 6}{/tex}    and 2=2×{tex}{6\over6}{/tex}

=> 1={tex}{6\over6}{/tex}   and 2={tex}{12\over 6}{/tex} 

So five rational numbers between 1={tex}{6\over 6}{/tex}and 2 ={tex}{12\over 6}{/tex} are 

{tex}{7\over 6}{/tex},{tex}{8\over 6}{/tex},{tex}{9\over 6}{/tex},{tex}{10\over 6}{/tex} and {tex}{11\over 6}{/tex}

In general if need to find n rational numbers between  a and b then we multiply and divide both the numbers by n+1, then write numbers from a(n+1) to b(n+1) and after that divide it by (n+1) we get required rational numbers between a and b.

in above problem a=1 b=2  and n=5 so we multiply and  divide 1 and 2 by 5+1=6.

Now the numbers between a(n+1)=1×6=6 to b(n+1)=2×6=12 are 7,8,9,10 and 11.

So required rational numbers are 7/6,8/6,9/6,10/6 and 11/6.

We can write 1 as {tex}{1 \over 1}{/tex} and 2 as {tex}{2 \over 1}{/tex}

Multiplying denominator by 10, we have

{tex}{1 \over 10}{/tex} and {tex}{2 \over 10}{/tex}

Now, you can choose any five rational numbers between 

{tex}{1 \over 10}{/tex} and {tex}{2 \over 10}{/tex}

{tex}{2 \over {10}},{3 \over {10}},{4 \over {10}},{5 \over {10}},{6 \over {10}},{7 \over {10}},{8 \over {10}},{9 \over {10}}{/tex}

Let a = 1 , b = 2 , n = 5 ,

b - a = 1 , n + 1 = 6

Therefore 5 rational numbers between 1 and 2 will be

a + (b - a) / 6 , a + 2(b - a) / 6, a + 3(b - a) / 6 , a + 4(b - a) / 6 , a + 5(b - a)/ 6

1 + 1/6 , 1 + 2/6 , 1 + 3/6 , 1 + 4/6 , 1 + 5/6 , 

7/6 , 8/6 , 9 / 6 , 10 / 6 , 11 / 6

Hence required rational numbers are 

7/6, 4/3, 3/2, 5/3, 11/6

please continue from above

3b = 180 - 15 = 165          b = 55

a = 55 - 30 = 25

c = 55 + 45 = 100

therefore  a = 25 , b = 55  ,  c = 100

In a triangle, we know that a + b + c = 180               ...............(i)

Given: b - a = 30          =>          b = a + 30       ..............(ii)

Also,    c - b = 45         =>           b = c - 45        ..............(iii)

Putting value of b from eq.(ii), in eq.(i), we get

a + a + 30 + c = 180

=>          2a + c = 150                ...............(iv)

Again putting the value of b from eq.(iii), in eq.(i), we get

a + c - 45 + c = 180

=>          a + 2c = 225               ................(v)

On solving eq.(iv) and eq.(v), we get

a = 25 and c = 100

Then, b = 30 + a = 30 + 25 = 55

Therefore, a = 25, b = 55, c = 100

b - a = 30         a =  b - 30   ....... i

c - b =45         c = b + 45 ..........ii

a + b + c = 180.....iii

substitute i and ii  in iii 

b - 30 + b + b + 45 = 180

ASA      Angle Side Angle     Rule 

Two triangles are congruent if the two angles and the included side are equal

similarly there are 5 ways of finding if the two tringles are congruent

SSS, SAS , AAS, ASA, HL

Caution   AAA       Angle Angle Angle

Triangles which have three equal angles , we cannot be sure if they are congruent without knowing at least one side    

AAS Congruence Criterion: If any two angles and a non-included side of one triangle are equal to the corresponding angles and side of another triangle, then the two triangles are congruent.

ASA is the Angle Side Angle rule.

In which we can say triangles are congruent by seeing their two angles and one side.

Using Pythagoras Theorm

{tex}Height = \sqrt {(hypotenuse)^2-(base)^2}{/tex}

{tex}= \sqrt{625-49}{/tex}

{tex}= \sqrt{576}= 24 {/tex} cm

Area of ∆ = {tex}{1\over 2}\times base\times height{/tex}

= {tex}{1\over2}\times 7\times 24= 84\ cm^2{/tex}

{tex}(0.001)^{1\over 3}{/tex}

= {tex}((0.1)^3)^{1\over 3}{/tex}

= {tex}(0.1)^{3\times {1\over 3}}{/tex}

= {tex}(0.1)^1= 0.1 {/tex}

{tex}(0.001)^{1\over 3}{/tex}

= {tex}((0.1)^3)^{1\over 3}{/tex}

= {tex}(0.1)^{3\times {1\over 3}}{/tex}

= {tex}(0.1)^1= 0.1 {/tex}

√12×√5=√60

             =2√15