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ANSWER

Show that 3 under root 6 is an irrational

Posted by Ansh Wadhwa 6 years, 7 months ago

CBSE > Class 10 > Mathematics

1 answers

Manish Pandhyar 6 years, 7 months ago

Let 3 under root 6 be a rational no. 3 under root 6 = P/Q where p and q are co-prime numbers and Q is not equal to 0 3 under root 6= PQ Under root6=P/3Q Root 6 is irrational no So p/3q is also irrational no So there is contradiction 3 under root 6 is irrational number

ANSWER

If alpha and beta are the zeroes of the polynomial f(x) =2x^2-5x+k such that (alpha +beta)^2-alpha*beta. =24 find k

Posted by Ashish Singh 6 years, 7 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 7 months ago

The given quadratic polynomial is 2x²+ 5x + k.
If {tex}\alpha, \beta{/tex} are zeroes of quadratic polynomial
{tex}\alpha + \beta= \frac { - b } { a } = \frac { - 5 } { 2 } {/tex}
{tex} \alpha \beta= \frac { c } { a } = \frac { k } { 2 }{/tex}
Putting these values in {tex}( \alpha + \beta ) ^ { 2 } - \alpha \beta{/tex} = 24,
we get {tex}\left( \frac { - 5 } { 2 } \right) ^ { 2 } - \frac { k } { 2 } = 24 {/tex}
{tex}\Rightarrow \frac { 25 } { 4 } - \frac { k } { 2 } = 24{/tex}

{tex}\Rightarrow\frac { - k } { 2 } = 24 - \frac { 25 } { 4 }{/tex}
{tex}\Rightarrow \frac { - k } { 2 } = \frac { 96 - 25 } { 4 }{/tex}
{tex}\Rightarrow k = \frac { - 71 } { 4 } \times 2 = \frac { - 71 } { 2 }{/tex}

ANSWER

Find HCF. Of 867&255

Posted by Abhi Khatri 6 years, 7 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 7 months ago

On applying division lemma to 867 and 255

We get

867=255×3+102

255=102×2+51

102=51×2+0

Hence HCF(867,255)=51

ANSWER

WHEN WILL SOLUTION OF RD SHARMA WILL COME FOR ALL CHAPTER....

Posted by Apurva Gholve 6 years, 7 months ago

CBSE > Class 10 > Mathematics

2 answers

Harsh Arora 6 years, 7 months ago

Today

Harsh Arora 6 years, 7 months ago

Tosay

ANSWER

(1- COS SQUARE A)cosec square a

Posted by Piyush Bhatia 6 years, 7 months ago

CBSE > Class 10 > Mathematics

1 answers

Pranav Ramesh 6 years, 7 months ago

[1-cos square a]cosec a (Sin square a)cosec a (1÷cosec square a)cosec a (1÷cosec a) (Sin a

ANSWER

2x+3y=4;(k+2)x+6y=3k+2 find the value of k if there exists unique solution

Posted by Bhavana Reddy 6 years, 7 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 7 months ago

{tex}2x + 3y - 4 = 0{/tex}

{tex}a_1= 2, b_1= 3, c_1 = 4{/tex}

{tex}(k + 2)x + 6y-(3k + 2)= 0{/tex}
{tex}a _ { 2 } = k + 2 , b _ { 2 } = 6 , c _ { 2 } = - ( 3 k + 2 ){/tex}

{tex}\frac { a _ { 1 } } { a _ { 2 } } = \frac { b _ { 1 } } { b _ { 2 } } = \frac { c _ { 1 } } { c _ { 2 } }{/tex}

or, {tex}\frac { 2 } { k + 2 } = \frac { 3 } { 6 } = \frac { 4 } {- 3 k - 2 }{/tex}
Or, {tex}\frac { 2 } { k + 2 } = \frac { 3 } { 6 } {/tex}

or, {tex}3(k + 2) = 2 \times 6{/tex}

{tex}3k + 6 = 12{/tex}
{tex}3k = 12 - 6{/tex}

or, {tex}6 = 3k{/tex}

Hence,{tex} k = 2{/tex}

ANSWER

Area of 4walls

Posted by Sandeep Sharma 6 years, 7 months ago

CBSE > Class 10 > Mathematics

1 answers

????? ? 6 years, 7 months ago

★If the 4 walls are square in shape, then find out the L.S.A. of cube=4a². ★If the 4 walls are rectangular in shape, then find out the L.S.A. of cuboid=2h(l+b).

ANSWER

What is the difference between chord and secant..? Frnds.. Plz tell me..

Posted by Anmol Kumar Jha 6 years, 7 months ago

CBSE > Class 10 > Mathematics

1 answers

????? ? 6 years, 7 months ago

A line segment which touches the circle at two points is called chord whereas a line which passes intersecting the circle at two points is called secant.

ANSWER

For any positive integer n, prove that n^3 - n is divisible by 6.

Posted by Raj Sekhar Paul 6 years, 7 months ago

CBSE > Class 10 > Mathematics

2 answers

Harsh Arora 6 years, 7 months ago

Value put karke dekhni padegi

Khushi Agrawal 6 years, 7 months ago

Bhai mere ko bhi eska answer jaanna hai agar pataa chal jaye too mere ko bhi bataa dena please......

ANSWER

4/3 {sin square 59 -cot ^31}- 2/3 sin 90+3 tan^56 tan^2 34=x/3

Posted by Dhairya Baranwal 6 years, 7 months ago

CBSE > Class 10 > Mathematics

1 answers

Anitadhy Mali 6 years, 7 months ago

21/3. Ha xnxn

ANSWER

Find largest no. which when divides 705 & 1053 leaving remainder 9 in each case

Posted by Sparsh Aggarwal 6 years, 7 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

Cot-1 (√1+sinx+√1-sinx/√1+sinx-√1-sinx)

Posted by Sandeep Singh 6 years, 7 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

If theeta is 90°

Posted by Piyush Saroj 6 years, 7 months ago

CBSE > Class 10 > Mathematics

3 answers

Khushi Agrawal 6 years, 7 months ago

Bhai poraa ques.likho ye complete question nahi hai

Kapil Pandey 6 years, 7 months ago

Pritam Rakshit 6 years, 7 months ago

Is zero is prime number

ANSWER

Please answer my question...

Posted by Suyash Kumar 6 years, 7 months ago

CBSE > Class 10 > Mathematics

1 answers

Khushi Agrawal 6 years, 7 months ago

Bhai aapka question kya hai please tell me☺?

ANSWER

The HCF of a^m + b^mand a^n - b^n,where m is a odd positive no. and n is a even positive integer is............

Posted by Suyash Kumar 6 years, 7 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

(1+cotA+tanA)(sinA - cosA)

Posted by Shrabana Santra 6 years, 7 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 7 months ago

L.H.S
= (1 + cotA + tanA) (sinA - cosA)
= sinA - cosA + cotA sinA - cotA cosA + sinA tanA - tanA cosA
{tex}= \sin A - \cos A + \frac{{\cos A}}{{\sin A}} \times \sin A - \cot A\cos A + \sin A\;\tan A - \frac{{\sin A}}{{\cos A}} \times \cos A{/tex}
= sinA - cosA + cosA - cotA cosA + sinA tanA - sinA
= sinA tanA - cotA cosA........(1)
Now taking ;
{tex}\quad \frac{{\sec A}}{{\cos e{c^2}A}} - \frac{{\cos ecA}}{{{{\sec }^2}A}}{/tex}
{tex} = \frac{{\frac{1}{{\cos A}}}}{{\frac{1}{{{{\sin }^2}A}}}} - \frac{{\frac{1}{{\sin A}}}}{{\frac{1}{{{{\cos }^2}A}}}}{/tex}
{tex} = \frac{{{{\sin }^2}A}}{{\cos A}} - \frac{{{{\cos }^2}A}}{{\sin A}}{/tex}
{tex} = \sin A \times \frac{{\sin A}}{{\cos A}} - \cos A \times \frac{{\cos A}}{{\sin A}}{/tex}
{tex} = \sin A \times \tan A - \cos A \times \cot A{/tex}.......(2)
From (1) & (2),
(1 + cotA + tanA) (sinA - cosA) = {tex}\frac { \sec A } { cosec ^ { 2 } A } - \frac { cosec A } { \sec ^ { 2 } A }{/tex} = sinA.tanA - cosA.cotA
Hence, Proved.

ANSWER

Find the value of k in 2x + 3y=7 and (k-1)x + (k+2)y=3k

Posted by Ayushi Tiwari 6 years, 7 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 7 months ago

{tex}2x + 3y = 7{/tex}
{tex}(k - 1) x + (k + 2)y = 3k{/tex}
These are of the form
{tex}a_1x + b_1y + c_1 = 0 ,\ a_2x + b_2y + c_2 = 0{/tex}
where,
{tex}a_1= 2 ,\ b_1= 3,\ c_1 = -7,{/tex}
{tex}a_2=k - 1 \ ,b_2= k + 2 ,\ c_2 = -3k {/tex} for infinitely many solutions, we must have
{tex}\frac { a _ { 1 } } { a _ { 2 } } = \frac { b _ { 1 } } { b _ { 2 } } = \frac { c _ { 1 } } { c _ { 2 } }{/tex}
This hold only when
{tex}\frac { 2 } { k - 1 } = \frac { 3 } { k + 2 } = \frac { - 7 } { - 3 k }{/tex}
{tex}\frac { 2 } { k - 1 } = \frac { 3 } { k + 2 } = \frac { 7 } { 3 k }{/tex}
Now the following cases arises:
Case I:
{tex}\frac { 2 } { k - 1 } = \frac { 3 } { k + 2 }{/tex}
{tex}\Rightarrow{/tex}2(k + 2) = 3(k -1)

{tex}\Rightarrow{/tex}2k + 4= 3k - 3
{tex}\Rightarrow{/tex} k = 7
Case II:
{tex}\frac { 3 } { k + 2 } = \frac { 7 } { 3 k }{/tex}
{tex}\Rightarrow{/tex}7(k + 2) = 9k

{tex}\Rightarrow{/tex}7k + 14= 9k
{tex}\Rightarrow{/tex} k = 7
Case III:
{tex}\frac { 2 } { k - 1 } = \frac { 7 } { 3 k }{/tex}
{tex}\Rightarrow{/tex} 7k - 7 = 6k
{tex}\Rightarrow{/tex} k = 7
For k = 7, there are infinitely many solutions of the given system of equations.

ANSWER

Find the value of k for which the given equation have real and equal roots :x²-2(k+1)x+k²

Posted by Tanuj Kumar 6 years, 7 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 7 months ago

{tex}k = \frac { - 1 } { 2 }{/tex}

ANSWER

Dosto?i m in 10 atd batvh of 2019-2020 , plz tell me how to do study of maths ????R.D sharma me jo questions vo thode tough ja rahe hai,ncert me us type k questions nhi diye hai . Tell me kya kru . Im getting pannic

Posted by Yash Bastawale,? 6 years, 7 months ago

CBSE > Class 10 > Mathematics

9 answers

Yash Bastawale,? 6 years, 7 months ago

Hey tell me, i m in 10 nd now my summer vacations is going?so how to utilize this time . I wnt to score upto90%. Give me some advice ,

Yash Bastawale,? 6 years, 7 months ago

Nd my name is yash

Yash Bastawale,? 6 years, 7 months ago

Yes gust 2 days have passed to download this app

Kishu ? 6 years, 7 months ago

Are you new in this app?

Kishu ? 6 years, 7 months ago

Yes ritesh

Yash Bastawale,? 6 years, 7 months ago

Hey , kishu . Have u crossed u r board examinations of 10

Kishu ? 6 years, 7 months ago

Yash Bastawale,? 6 years, 7 months ago

?thanks ,

Kishu ? 6 years, 7 months ago

Dekhiye ap panic mat hoiye...ap pehle acche se ncert kariye...aur haan rd sharma me bohat saare concepts extra bhi hai jo 10th level ke nahi hai...jitna ncert me hai utna basics practice kariye...aur sample papers solve karo jyada sd jyada...isse apko paper ka level pata chalega..and dnt be panic..just chill...?hope it helps u out

4Thank You

ANSWER

Solve for x and y 1\(3x+y)+1\(3x-y)=3\4

Posted by Bijendra Kisku 6 years, 7 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

Find a quadratic polynomial whose zero are 4+√5 and 4-√5

Posted by Shreya Chand 6 years, 7 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 7 months ago

Let α=4 - {tex}\sqrt 5{/tex}
and β = 4 + {tex}\sqrt 5{/tex}
Sum of zeroes,
α+β = 4 - {tex}\sqrt 5{/tex} + 4 + {tex}\sqrt 5{/tex} = 8
αβ = (4 - {tex}\sqrt 5{/tex})(4 + {tex}\sqrt 5{/tex}) = 16 - 5 = 11
Quadratic polynomial
p(x) = x² - (α+β)x +αβ
= x² - 8x + 11