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Hlw hlw ....koi h ...itthe ...
  • 13 answers

Jaani Kaur🔥 7 hours ago

Same pitch😉
32℃ h ... Ab smj jaao ... Baarish ... Aandi .... Or aapke yaha

Jaani Kaur🔥 7 hours ago

Howz the weather there???
Accha

Jaani Kaur🔥 7 hours ago

PUNJAB🔥
Haha ....m mumbai se hu or aap

Jaani Kaur🔥 7 hours ago

Apne bluish room se😂🔥
Ghr se or aap

Jaani Kaur🔥 8 hours ago

You are from.......?
Im also guuud

Jaani Kaur🔥 8 hours ago

Gud U say????
Hii jaani ....hru ?

Jaani Kaur🔥 8 hours ago

Yea.....
Use Euclids division lemma to show that the cube of any positive integer of the form 9m or 9m+1 or 9m+8
  • 1 answers

Jaani Kaur🔥 8 hours ago

Let "a" be any positive integer and 9 be another . As we know that every positive integer can be written in the form of 3q , 3q+1 , 3q+2 Also , a=9q+r where 0 is equal to or is lower than r(remainder) and 0 < 9....by euclid divison lemma Now , put r = 0 a=9q a^3 = (9q)^3 = 729q^3 = 9(81q^3) = 9m where m=81q^3 Put r=1 a=9q+1 a^3 =(9q+1)^3 = 729q^3 + 1 + 243q^2 +27q = 9(81q^3+27q^2+3q) +1 =9m+1 where m =81q^3+27q^2+3q Put r=2 a=9q+2 a^3=(9q+2)^3 =729q^3 +8+486q^2 +108q = 9(81q^3+54q^2+12q)+8 = 9m+8 where m=81q^3 +54q^2+12q Hence proved .
Plz answer my ques
  • 1 answers

Pranav Mittal an hour ago

What is the question
If polynomial 6x4 + 8 x cube + 17 X square + 21 X + 7 is divided by another polynomial 3 X square + 4 x + 1 the remainder comes out be bracket a x + b bracket close then find a and b
  • 1 answers
Firstly divide the polynomial by 3x^2 + 4x+ 1 then the remainder u will get is X + 2 A/q, ax + b = x + 2 Hence , a = 1 and b = 2.
Find k if sum of zeros of polynomial x²-(k+6)x+2(2k-1) is half their product
  • 1 answers
I think K = 7. A/q, k + 6 = 1/2 × 4k - 2 k + 6 = 2k - 1 k - 2k = - 1- 6 K = 7.
Find the largest number that will divide 398 436 and 542 leaving remainders 7 11 and 15 respectively
  • 1 answers
The required number = HCF ( 398-7),(436-11),(542-15) = HCF (391),(425),(527) = 17.
Form the cubic polynomial whose zeros are minus 3 minus 1 and 2
  • 1 answers
x^3 -(-3)x^2 + (-1)x - 2 = x^3 + 3x^2 - x - 2
Find the zeroes of f (x)=3x^2+11x-4
  • 1 answers

Soni Suraj 13 hours ago

3x^2+11x-4 3x^2+12x-1x-4 3x(x+4)-1(x+4) (3x-1)(x+4) x= 1/3 , -4
What is quardatic equation
  • 3 answers
Quadratic equeation is those equation when the power of equation is two

Gaurav Seth 17 hours ago

QUADRATIC EQUATIONS

The polynomial of degree two is called quadratic polynomial and equation corresponding to a quadratic polynomial P(x) is called a quadratic equation in variable x.

Thus, P(x) = ax2 + bx + c =0, a ≠ 0, a, b, c ∈ R is known as the standard form of quadratic equation.

There are two types of quadratic equation.
(i) Complete quadratic equation : The equation ax2 + bx + c 0 where a ≠ 0, b ≠ 0,c ≠ 0
(ii) Pure quadratic equation : An equation in the form of ax2 = 0, a ≠ 0, b = 0, c = 0

Tushar Singh 18 hours ago

A equation whose maximum power is 2 called quadratice equation
If n is an odd integer then show n²-1is divisible by 8
  • 1 answers

Sia 🤖 16 hours ago

Let n = 4q + 1 (an odd integer)
{tex}\therefore \quad n ^ { 2 } - 1 = ( 4 q + 1 ) ^ { 2 } - 1{/tex}
{tex}= 16 q ^ { 2 } + 1 + 8 q - 1 \quad \text { Using Identity } ( a + b ) ^ { 2 } = a ^ { 2 } + 2 a b + b ^ { 2 }{/tex}
{tex}= 16{q^2} + 8q{/tex}
{tex}= 8 \left( 2 q ^ { 2 } + q \right){/tex}
= 8m, which is divisible by 8.

Find the root of 2x-3/x-2+2x-7/x-4=16/3
  • 1 answers

Rajan Kumar Pasi 1 day, 1 hour ago

{tex}Question:\space\large{ {2x-3\over x-2} +{2x-7\over x-4} = {16\over 3}, find \space x?}{/tex}

{tex}\large {Solution:\\ } {/tex}

{tex}\large { \underline {Step\space 1:\space Fraction\space solving} } {/tex}

{tex}\large\implies {{\big((2x-3)(x-4)\big)} +{\big((2x-7)(x-2)\big)}\over {(x-2)(x-4)}} = {16\over 3}{/tex}

{tex}\large\implies {{\big((2x^2-8x-3x+12)\big)} +{\big((2x^2-4x-7x+14)\big)}\over {(x^2-4x-2x+8)}} = {16\over 3}{/tex}

{tex}\large\implies {{\big(2x^2-11x+12+2x^2-11x+14\big)}\over {(x^2-6x+8)}} = {16\over 3}{/tex}

{tex}\large\implies {{\big(4x^2-22x+26\big)}\over {(x^2-6x+8)}} = {16\over 3}{/tex}

{tex}\large{ \underline {Step\space 2:\space Cross-multiplication}\\\implies {{3\times \big (4x^2-22x+26\big)} } = {16\times {(x^2-6x+8)}}}{/tex}

 

{tex}\large{ \underline {Step\space 3:\space Simplification:}\\\implies {{\big (12x^2-66x+78\big)} } = {{(16x^2-96x+128)}}}{/tex}

 

{tex}\large {\underline {Step\space 4:\space Equation-solving:}\\\implies 16x^2-96x+128-12x^2+66x-78=0}{/tex}

{tex}\large {\underline {Step\space 5:\space Arranging\space like-terms:}\\\implies 16x^2-12x^2-96x+66x+128-78=0}{/tex}

{tex}\large{\\\implies 4x^2-30x+50=0}{/tex}

 

{tex}\large {\underline {Step\space 6:\space Taking\space out\space common-terms:}\\\implies 2\times (2x^2-15x+25)=0}{/tex}

{tex}\large {\\\implies 2x^2-15x+25=0}{/tex}

{tex}\large {\underline {Step\space 7:\space Middle-term\space splitting\space :}\\\implies 2x^2-10x-5x+25=0}{/tex}

{tex}\large{ \underline {Step\space 8:\space Grouping\space :}\\\implies 2x(x-5)-5(x-5)=0}{/tex}

{tex} \\\implies \large(2x-5)(x-5)=0{/tex}

{tex} \\\implies\huge \therefore \boxed{x=5\space or \space {5\over2}}{/tex}





 

Solve : 2/x+3/3y =1/6 and 3/x+2/y =0
  • 2 answers

Saraswati Sabat 18 hours ago

3(2/x +3/3y = 1/6) 2(3/x+2/y = 0) 6/x +3/y = 1/2 6/x +4/y = 0 (-) (-) 0 + (-y) = 1/2 y = -2 2/x - 3/6 = 1/6 2/x = 1/6 + 3/6 2/x = 4/6 4x = 12 x = 3

Rajan Kumar Pasi 1 day, 1 hour ago

This is a very simple question of linear equations in two variables chapter, I think you have to practice more on fractions arithmetic

{tex}\large Question:\frac 2x +\frac 3{3y}=\frac 16\space and\space \frac 3x +\frac 2y=0,\space find\space x,y?{/tex}

Solution:

<th colspan="2" scope="col">

{tex}\large\implies\boxed {\frac {2\times 3y +3\times x}{x\times 3y}=\frac 16 \space \dots eq.(i)}{/tex}

{tex}\large{\implies {\frac {6y +3x}{3xy}=\frac 16 \space }}{/tex}

{tex}\large {\underline {Step\space 2:\space Cross-multiplication:} }{/tex}

{tex}\large{\implies {6\times{(6y +3x)}=1\times {3xy} \space }}{/tex}

{tex}\large{\implies {36y +18x}={3xy} \space }{/tex}

{tex}\large{\implies {36y +18x}-{3xy}=0 \space }{/tex}
 

{tex}\large {\implies\boxed {\frac {3\times y +2\times x}{x\times y}=0 \space\dots eq.(ii) }}{/tex}

{tex}\large {\implies\frac {3y +2x}{xy}=0 \space }{/tex}

{tex}\large {\underline {Step\space 2:\space Cross-multiplication:} }{/tex}

{tex}\large {\implies{3y +2x}=0\times {xy}=0 \space }{/tex}

{tex}\large {\implies\boxed {{3y}=-2x \space \dots eq.(iii)}}{/tex}

 

 

{tex}\large {\underline {Step\space 3:\space Putting\space relation\space of\space eq.(iii):} }{/tex}

{tex}\large{\implies {{\underline {3y}}(12-x) +18x}=0 \space }{/tex}

{tex}\large{\implies {\underline{(-2x)}(12-x) +18x}=0 \space }{/tex}

{tex}\large{\implies {-24x+2x^2 +18x}=0 \space }{/tex}

{tex}\large{\implies {2x^2-6x}=0 \space }{/tex}

{tex}\large{\implies {2x^2=6x \space} }{/tex}

{tex}\large{\implies\boxed {x=2 \space} }{/tex}

{tex}\large {\underline {Step\space 4:\space Putting\space value\space of\space 'x'\space in\space eq.(iii):} }{/tex}{tex}\large {\implies{3y}=-2x \space }\\ \large {\implies{3y}=-2\times 2 \space }\\ \large\implies\boxed {{y}={-4\over3} \space }\\{/tex}

 

Prove √2+5√2 ....be irrational.. .....(plz koi ans de dena plz plz )
  • 12 answers

Kiara Khurrana 20 hours ago

yes your mother and father is thinking that my son and daughter is studing but how shameful it is to see these things

Diya 1 day, 6 hours ago

Mujhe lga tum chle gye phir se....😅

Rgr 😋 1 day, 6 hours ago

Ro kyu rahi ho Phir?

Rgr 😋 1 day, 6 hours ago

Hmm

Diya 1 day, 6 hours ago

😭😭

Diya 1 day, 6 hours ago

Nhii hu naraj.😊😚😊😊😉..teri yaar hu mai😄😄😄yhi song h na...😉

Rgr 😋 1 day, 6 hours ago

Ans to do Diya 😭😭😞😞

Rgr 😋 1 day, 7 hours ago

Abhi tak naraz ho na hamse, Diya 😞😞😞?

Diya 1 day, 7 hours ago

I hope this will help u bestie😘😘😄

Rgr 😋 1 day, 7 hours ago

Tu jo rutha to kon hasega.

Diya 1 day, 7 hours ago

This ques. is v. simple bestie. 😘First of all iss mese common le lo ✓2...that is ✓2(1+5)=6✓2 and now we have to prove that 6✓2 is irrational. Let 6✓2 is a rational no. Therefore, 6✓2= p/q(where p and q are co-prime) ✓2=p/6q, p/6q is rational(as p and q are integers) therefore, ✓2 is rational. But this contradicts the fact that ✓2 is irrational. Therefore, 6✓2 is irrational .😊😊

Rgr 😋 1 day, 7 hours ago

Root 2 ko jase irrational prove karte hai whsehi karo
Find the quadratic polynomial whose zeroes are 3+ √5 and 3_√5
  • 1 answers

This also seems like you didn't learn chapter 4 in a better way.

{tex}\large{Question: \space Zeroes=3+\sqrt5 \space and \space 3-\sqrt 5,\space find\space the \space quadratic\space equation?}{/tex}

{tex}\large Solution:{/tex}

{tex}\large {For\space any\space quadratic\space equation,\space x^2+x\times(\alpha+\beta)+(\alpha\times\beta)=0\space }{/tex}

{tex}\large {Here,\space let\space\space\alpha=3+\sqrt5 \space\space\space\space,\space\space\space\space\beta=3-\sqrt5.\space }{/tex}

{tex}\large {Thus,\space\space\space\alpha+\beta=3+\sqrt5 +3-\sqrt5 = 6\space\space\space\space,\space\space and\space\space\alpha\times\beta=(3+\sqrt5)(3-\sqrt5)=3^2-\sqrt5 ^2=9-5=4.\space }{/tex}

{tex}\large {\therefore\space x^2+x\times(\alpha+\beta)+(\alpha\times\beta)=0\space }\\ \large {\implies\space x^2+x\times(6)+(4)=0\space }\\ \large {\implies\boxed{\space x^2+6x+4=0}\space }\\{/tex}

Now after seeing this, don't you think this is a very very simple thing to do.

Can 72 and 20 be the L.C.M and H.C.F of two numbers? Write down the reason.
  • 1 answers

Question : Can 72 and 20 be the L.C.M and H.C.F of two numbers? Write down the reason.

This is a very pathetic question to ask. It really shows that you are not studying well.

Solution: {tex}\large{\implies \boxed{LCM\times HCF=a\times b}}{/tex}

{tex}\large{\implies{72\times20=a\times b}}{/tex}                    {tex}\large{\implies{1440=a\times b}}{/tex}

Now, 1440 can be multiple of any two numbers.

{tex}\large{\implies{\space\space1\times1440=1440 }}\\ \large{\implies{\space\space2\times720\space\space=1440 }}\\ \large{\implies{\space\space3\times480\space\space=1440 }}\\ \large{\implies{\space\space4\times360\space\space=1440}}\\ \large{\implies{\space\space5\times288\space\space=1440}}\\ \large{\implies{\space\space6\times240\space\space=1440}}\\ \large{\implies{\space\space8\times180\space\space=1440}}\\ \large{\implies{\space\space9\times160\space\space=1440 }}\\ \large{\implies{10\times144\space\space=1440 }}\\ \large{\implies{12\times120\space\space=1440 }}\\ \large{\implies{15\times96\space\space\space\space=1440 }}\\ \large{\implies{16\times90\space\space\space\space=1440 }}\\ \large{\implies{18\times80\space\space\space\space=1440 }}\\ \large{\implies{20\times72\space\space\space\space=1440 }}\\ \large{\implies{24\times60\space\space\space\space=1440 }}\\ \large{\implies{30\times48\space\space\space\space=1440 }}\\ \large{\implies{32\times45\space\space\space\space=1440 }}\\ \large{\implies{36\times40\space\space\space\space=1440 }}\\ {/tex}

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