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  • 1 answers

Ashnoor Kaur 6 years, 5 months ago

Please write the question correctly
  • 0 answers
  • 2 answers

Gaurav Tanwar 6 years, 5 months ago

X-49 YOUR QUESTION IS NOT COMPLETE?

Unknown Girl 6 years, 5 months ago

x-54+5=0 => x=54-5 =>x=14 Hope ii will help u
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  • 0 answers
  • 1 answers

Sia ? 6 years, 5 months ago

159

  • 0 answers
  • 1 answers

Rekha Nayak 6 years, 5 months ago

TanA+SecA-1/TanA-SecA+1=1+SinA/CosA
  • 1 answers

Sia ? 6 years, 5 months ago

One zero {tex}= -5{/tex}
product of zeroes {tex}= 0{/tex}
{tex}\therefore{/tex} Other zero = {tex}\frac { 0 } { - 5 }{/tex} {tex}= 0{/tex}
Sum of zeroes {tex}= -5 + 0 = -5{/tex}
Polynomial {tex}p(x) = x^2 - (S)x + P{/tex}
= {tex}x^2 + 5x{/tex}

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Sagar Av 6 years, 5 months ago

25x^2 - 15x +2 =0 In this equation a=25,b=-15,c=2 a+b=-b/a a*b=c/a a+b= -(-15)/25 a*b= 2/25 Sum= 1/a+1/b b+a/ab 15/25/2/25 3/5x25/2 = 15/2 Product = 1/a *1/b 1/ab P=25/2 ax^2+bx+c=0 25x^2+15/2+25/2 50x^2+15x+25=0 Divide the equation by 5 10x^2+3x+5=0 The equation is 10x^2+3x+5
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Ashish Parcha 6 years, 5 months ago

From 1 eq.. X+2y=3 X=-2y-3 2...... -2x-y=4 -2(2y-3)-y-4=0 -4y-6-y-4=0 5y-10=0 y=10/5 y=2 Puting in third X=-2y-3 X=-2(2)-3 X=-4-3 X=-7
  • 1 answers

Sia ? 6 years, 5 months ago

  1. In DAPB and DAQB
    {tex}\angle{/tex}BAP = {tex}\angle{/tex}BAQ ...[As l bisects {tex}\angle{/tex}A]
    {tex}\angle{/tex}BPA = {tex}\angle{/tex}BQA ...[Each 90°]
    AB = AB ...[Common]
    {tex}\therefore{/tex} DAPB {tex}\cong{/tex} DAQB proved ...[AAS property] ...(1)
  2. DAPB {tex}\cong{/tex} DAQB ...[From (1)]
    {tex}\therefore{/tex} BP = BQ ...[c.p.c.t.]
  • 1 answers

Nm ???? 6 years, 5 months ago

This question is incomplete
  • 5 answers

Lina Choudhary 6 years, 5 months ago

trigonometry

Pramod Kumar 6 years, 5 months ago

Trigonometry

Reena Jain 6 years, 5 months ago

Trigonometry

Nm ???? 6 years, 5 months ago

Ch.15,14 ,1,2,5,3 the and tenometry are the most important chapters

Account Deleted 6 years, 5 months ago

Trigonometry i guess
  • 1 answers

N Pranav Tej 6 years, 5 months ago

x^2+6x+8=0 x^2+2x+4x+8 x(x+2)+4(x+2) (x+2)(x+4) therefore -2&-4 are zeros of the polynomial
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Nm ???? 6 years, 5 months ago

We have here 43,91,183 So the difference r 183-91=92 183-43=140 91-43=48 Now Hcf(48,140,92) As 48=2×2×2×2×3 92=2×2×23 140=2×2×5×7 Hcf = 2×2×=4 And 4 is the required number

Komal Preet 6 years, 5 months ago

How to solve it

Aanchal ????? 6 years, 5 months ago

183
  • 4 answers

Account Deleted 6 years, 5 months ago

Well i guess your question is incomplete coz its not s class 10 th question at all

Mohammed Daanish 6 years, 5 months ago

-82

❤ 753 6 years, 5 months ago

Sure u r in class 10??

Manas Rathod 6 years, 5 months ago

-82
  • 3 answers

Nm ???? 6 years, 5 months ago

R u sure u r in 10th class because this nost the ques. Of 10th class

Mohammed Daanish 6 years, 5 months ago

77590

@ Aashu 6 years, 5 months ago

77590
  • 1 answers

N Pranav Tej 6 years, 5 months ago

60=2×2×3×5 = 2^2×3×5 72=2×2×2×3×3=2^3×3^2 HCF(60,72)=2^2×3=12 LCM(60,72)=2^3×3^2×5=360 Here LCM(60,72)×HCF(60,72)=60×72 LHS=LCM(60,72)×HCF(60,72) =360×12 =4,320 RHS=60×72 =4,320 Here RHS=LHS HENCE PROVED
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Nm ???? 6 years, 5 months ago

Plz tell full question What we have to find in this question
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  • 2 answers

Sia ? 6 years, 5 months ago

Let us assume that <m:omath><m:r>√8</m:r></m:omath> is rational.

That is, we can find integers a and b (<m:omath><m:r>≠0)</m:r></m:omath> such that a and b are co-prime

{tex}\style{font-family:Arial}{\begin{array}{l}\sqrt8=\frac ab\\b\sqrt8=a\\on\;squaring\;both\;sides\;we\;get\\8b^2=a^2\end{array}}{/tex}

Therefore, a2 is divisible by 8,

 it follows that a is also divisible by 8.

So, we can write a = 8c for some integer c.

Substituting for a, we get 8b2 = 64c2, that is, b​​​​​​2 = 8c2

This means that b2 is divisible by 8, and so b is also divisible by 8

 Therefore, a and b have at least 8 as a common factor.

But this contradicts the fact that a and b are co-prime.

This contradiction has arisen because of our incorrect assumption that <m:omath><m:r>√8</m:r></m:omath> is rational.

So, we conclude that <m:omath><m:r>√8</m:r></m:omath> is irrational.

Nm ???? 6 years, 5 months ago

Firsty let root 8 is rational Then root8/1=p/q(p and q r coprime anf q is not equal to 0) Then cross multiplication then we obtain P=root8q Square on both side P2=(root8q)2 Then we obtain P2=8q2 because root se square cancel ho jayga Then we can say that p2 divide 8exactly and p will also divide 8 exaclty Then by euclid division lemma P=8q+0 Then square on both side P2=(8q)2 Then put the value of p2 8q2=64q2 8q2=8(8q2) Then 8 se 8 cancel ho jayga Then q2=8q2 Then we can say that q2 will divide 8 exactly and q will divide exactly Then the common factor of p and p is 8 then it is contraduction to our supposition. So, our supposition is wrong . Hence root 8 is irrational no.

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