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Show 3 root 7 is an irrational no.

Posted by Rohan Kumar 7 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Vidhi Jain 7 years, 6 months ago

let us take on contrary that 3 root 7 is rational. then there exist two co prime numbers a and b such that 3root 7 = a/b => root 7 = a / 7b now LHS is an irrational no. whereas RHs is rational this contradiction has arised due to our wrong assumption in beginning therefore 3 root 7 is an irrational number

ANSWER

Prove that if X and y are odd positive integers, then x2 +y2 is even but not divisible by 4?

Posted by Gaurav Kaushik 7 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Susai Raj 7 years, 6 months ago

x and y are odd integers. So they should be of the form x= 2m+1, y= 2n+1 for some integers m and n.
x^2 + y^2 =(2m+1)^2 + (2n+1)^2
=4m^2+4m +1+4n^2+4n+1
=4m^2+4n^2+4m+4n+2
=4(m^2+n^2+m+n) + 2
Which is an even number but not divisible by 4.

ANSWER

Prove that tanA +sinA ÷ tanA-sinA = secA+ 1÷secA-1

Posted by Amit Jangra 7 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Susai Raj 7 years, 6 months ago

(tanA+sinA)/(tanA-sinA)
={(sinA/cosA) + sinA} / {(sinA/cosA) -sinA}
=sinA{(1/cosA) + 1} / sinA{(1/cosA)- sinA}
=(1/cosA + 1)/1/cosA - 1)
=(secA+1)/(secA-1).

ANSWER

Please send me maths lab manual activity of "Geometrical meaning of the zeroes of the polynomial".

Posted by Sayantika Roy 7 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Saurabh Jain 7 years, 6 months ago

Nice question

4Thank You

ANSWER

Write the following. Condition for linear equation having intersecting line for graphical representation

Posted by Sonal Gandhi 7 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Susai Raj 7 years, 6 months ago

If ax+by+cz=0 and
lx+my+n =0 , then if
a/l is not equal to b/m then the equations are consistent with a unique solution. In this case we will get intersecting lines in their graphical representation.

ANSWER

If a and b are the zeroes of a quadratic polynomial such that a+b=24and a-b=8. Find the quadratic polynomial representing a and b as its zeroes. Also verify the relation between the sum of its zeroes and product.

Posted by Cutiepie ❤ 7 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Susai Raj 7 years, 6 months ago

a+b=24.....(1)
a - b=8....(2)
We know that
(a+b)^2 - (a-b)^2 = 4ab
So 24^2 - 8^2 =4ab
ie (24+8)(24-8) =4ab
=> ab = 32×16/4
ie. ab = 128
So the required quadratic polynomial is
x^2-(sum of zeroes)x+ product of zeroes
ie. x^2 -24x +128
Sum of zeroes = -b/a =
-(-24 )/1 = 24
Product of zeroes = c/a
=128/1 = 128.

ANSWER

sin30°+tan45°-cosec60°÷sec30°+cos60°+cot45°

Posted by Ajay Manish 7 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Preyanshu Sinha 7 years, 6 months ago

Solution 1/2+1-2/root3÷ 2/root3+1/2+1 =root3+2root3-2root2/ 2root3. ÷ 2*2+root3+2root3/ 2root3 =root3+2root3-4÷ 4+root3+2root3 =(3root3-4)whole square÷ (3root3)whole square-4square =27+16-24root3÷ 27-16 =43-24rooot3÷ 11

3Thank You

ANSWER

Find the LCM and HCF of 625, 1125, 2125 using fundamental theorem of arithmetic

Posted by Prachi Shukla 7 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Susai Raj 7 years, 6 months ago

625 = 5^4
1125 =3^2 × 5^3
2125= 5^3 × 17
The required LCM =3^2×5^4×17 = 95625
The required HCF =5^3=125

ANSWER

Find the HCF of bracket start 2 power 3 multiply 3 power 3 multiply 5 bracket close text bracket s t a r t 2 power 2 multiply 3 power 3 x power 2 bracket close and Bracket s t a r t 2 ki power 4 multiply 3 multiply 5 ki power 3 multiply 7 Please answer quickly

Posted by Ravina Goyat 7 years, 6 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

If two positive integers P and Q can be expressed as B equal a cube b cube square and cube equal a b cube c square, A, B, C been prime numbers then what is the LCM of P and Q Please answer quickly

Posted by Ravina Goyat 7 years, 6 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

The LCMof two no.is 14time where HCF the sum of LCM andHCF is 600 if one no.is 280 then find the other number

Posted by Kirti Singh 7 years, 6 months ago

CBSE > Class 10 > Mathematics

2 answers

Ravina Goyat 7 years, 6 months ago

Let the HCF x and LCM 14x HCF plus LCM equal 600 X + 14 x = 615 x = 6 cm equal 40 HCF equal 40 and LCM equal 5 00 HCF multiply LCM first number multiply second number 40 multiplied 500 equals 28 multiplied second number second number second number equal to multiply 500 / 287 second number is 80

Ravina Goyat 7 years, 6 months ago

Given HCF + LCM=600 One number is 280

ANSWER

d²=1/4 how to be solve

Posted by Sanskar Bhardwaj 7 years, 6 months ago

CBSE > Class 10 > Mathematics

2 answers

Rishu Raj 7 years, 6 months ago

d\2=1/4 then u can write it as 4d\2=1 and after this u can write it as (2d)\2=1 or 2d=√1 or 2d=1 now d=1/2 this is the second process i explained ,here d\2means 2 power on d and 4d\2 means 4 multiply d square....

Shiva Yadav Shiva Yadav 7 years, 6 months ago

D=1/2

ANSWER

What is linear polynomial

Posted by Sonu Kumar 7 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Amr G 7 years, 6 months ago

A polynomial of degree 1 for eg x+ 1 here the highest power of the variable means degree is 1 so it is a linear polynomial

ANSWER

if one zero of the polynomial[ a2plus9]x2 plus 45x plus 6a find the value of a

Posted by Navjot Kaur 7 years, 6 months ago

CBSE > Class 10 > Mathematics

2 answers

Kannu Kranti Yadav 7 years, 6 months ago

Yaar,write ur que properly.

Navjot Kaur 7 years, 6 months ago

please answer quickly

ANSWER

Show that root 5+2root3 is irrational

Posted by Rita Raman 7 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Rishu Raj 7 years, 6 months ago

let 5+√3 =a is rational, then √3=a-5(1) we know that if rational is subtracted from rational we get rational as result ,so. from eq (1) irrational = rational ,which is impossible ,therefore our supposition is wrong ,and 5+√3=a is irrational.....

ANSWER

✓2x2+7x+5√2=0

Posted by Diwakar Sharma 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

We have, {tex}\sqrt2x^2 +7x +5\sqrt2 =0{/tex}

{tex}\implies \sqrt2x^2 +2x +5x+5\sqrt2 =0{/tex}

{tex}\implies \sqrt2x (x +\sqrt2) + 5 (x + \sqrt2) = 0{/tex}

{tex}\implies (\sqrt2x+5) (x+\sqrt2) =0{/tex}

{tex}Either\, \sqrt2x+5 =0\, or \,x+\sqrt2 =0{/tex}

{tex}\implies x = {-5 \over \sqrt2},\, -\sqrt2{/tex}

{tex}\therefore x = {-5 \over \sqrt2},\, -\sqrt2{/tex} are the required roots.

ANSWER

Surface Area and volume chapter all formula

Posted by Amanullah Khan 7 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Amanullah Khan 5 years, 8 months ago

Please answer

ANSWER

How understand the math

Posted by Bhabendra Kumar 7 years, 6 months ago

CBSE > Class 10 > Mathematics

2 answers

Rishu Raj 7 years, 6 months ago

Never take math as a big dodge or weight,always take it easy and play with math ,anytime u notice if u are playing with someone he also plays with u ofcourse he is your enemy or friend so if u play with math math will play with u and make itself easy to learn to u .......please take this point seriously u will become genius in math its my experience .....

Krituk Nikhra 7 years, 6 months ago

By concentration and by ur mind.... Take it as a song

ANSWER

How to learn the trigonomettic ratios

Posted by Manjesh Bhaskar 7 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Krituk Nikhra 7 years, 6 months ago

Only learn sinA, cosA and TanA Than u easily learn cosecA which is resiprocal of sinA, secA resiprocal of cosA and cotA resiprocal of tanA...

ANSWER

verify the relationship b/w the zeroes and coefficients[3x square plus 4x -4]

Posted by Navjot Kaur 7 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Rohit Kumar 7 years, 6 months ago

p(x)=3x^2+4x-4 here a=3,b=4 and c=-4 ( by comparison with ax^2+bx+c)
for x=0
p(x)=o
3x^2+4x-4=0
3x^2+6x-2x-4=0
3x(x+2)-2(x+2)=0
(3x-2)(x+2)=0

therefore x=2/3 or x=-2
let,
alpha=2/3
beta=-2

now, alpha+beta= -b/a
2/3-2=-4/3
-4/3=-4/3
similarly,
alpha x beta= c/a
2/3 x -2 = -4/3
-4/3 = -4/3

hence verified

4Thank You

ANSWER

a)Show that any positive odd integer is of the form 4q+1, 4q+3, where q is some integer.b) Show that any positive even integer is of the form 6q, 6q+2, 6q+4. Where q is some inteher

Posted by Kanika Kalsyan 7 years, 6 months ago

CBSE > Class 10 > Mathematics

2 answers

Navjot Kaur 7 years, 6 months ago

search on google

Kanika Kalsyan 7 years, 6 months ago

Please hurry up

ANSWER

The values of remainder r, when a positive integer a is divided by 3 are 0 and 1 only. Justify your answer.

Posted by Kanika Kalsyan 7 years, 6 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

In given figuer ABCD is a rectangle .length is 36. Cm and breadtg is 10 cm ..Find x and y

Posted by Bikramjit Singh 7 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Krituk Nikhra 7 years, 6 months ago

R u sure u had written as ques as given

ANSWER

. If Sin θ = m/n, find the value of tan θ + 4 / 4 cot θ +1.

Posted by Shakir Hussain 7 years, 6 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

Solve graphically the system of linear equation 2x-y=4 , x-y

Posted by Prabhjot Singh 7 years, 6 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

If alpha & beta r the zeroes of x2- 3x - 2 find a quadratic polynomial whose zeros r 1/2 alpha + beta & 1/2 beta + alpha

Posted by Vojal Harmukh 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

The given quadratic polynomial is
f(x) = x² - 3x - 2
Sum of the zeroes = α + β = 3
Product of the zeroes = αβ = -2
From the question, we have the quadratic polynomial ax² + bx + c whose zeroes are {tex}\frac{1}{{2\alpha + \beta }} \space and\ \space \frac{1}{{2\beta + \alpha }}{/tex}.
Sum of the zeroes of the new polynomial {tex}= \frac{1}{{2\alpha + \beta }} + \frac{1}{{2\beta + \alpha }}{/tex}
{tex} = \frac{{2\beta + \alpha + 2\alpha + \beta }}{{\left( {2\alpha + \beta } \right)\left( {2\beta + \alpha } \right)}}{/tex}
{tex}= \frac{{3\alpha + 3\beta }}{{2\left( {{\alpha ^2} + \beta^2} \right) + 5\alpha \beta }}{/tex}
{tex} = \frac{{3 \times 3}}{{2\left[ {{{\left( {\alpha + \beta } \right)}^2} - 2\alpha \beta + 5 \times \left( { - 2} \right)} \right]}}{/tex}
{tex}= \frac{9}{{2\left[ {9 - ( - 4)} \right] - 10}}{/tex}
{tex}= \frac{9}{{2\left[ {13} \right] - 10}}{/tex}
{tex} = \frac{9}{{26 - 10}} = \frac{9}{{16}}{/tex}
Also Product of the zeroes {tex}\frac{1}{{2\alpha + \beta }} \space \times\ \space \frac{1}{{2\beta + \alpha }}{/tex}
{tex}= \frac{1}{{\left( {2\alpha + \beta } \right)\left( {2\beta + \alpha } \right)}}{/tex}
{tex}= \frac{1}{{4\alpha \beta + 2{\alpha ^2} + 2{\beta ^2} + \alpha \beta }}{/tex}
{tex} = \frac{1}{{5\alpha \beta + 2\left( {{\alpha ^2} + {\beta ^2}} \right)}}{/tex}
{tex} = \frac{1}{{5\alpha \beta + 2\left( {{{\left( {\alpha + \beta } \right)}^2} - 2\alpha \beta } \right)}}{/tex}
{tex} = \frac{1}{{5 \times \left( { - 2} \right) + 2\left( {{{\left( 3 \right)}^2} - 2 \times \left( { - 2} \right)} \right)}}{/tex}
{tex}= \frac{1}{{ - 10 + 26}} = \frac{1}{{16}}{/tex}
So, the quadratic polynomial is,
x² - (sum of the zeroes)x + (product of the zeroes)
{tex}= \left( {{x^2} + \frac{9}{{16}}x + \frac{1}{{16}}} \right){/tex}
Hence, the required quadratic polynomial is {tex}\left( x ^ { 2 } + \frac { 9 } { 16 } x + \frac { 1 } { 16 } \right){/tex}.

ANSWER

Suppose you drop die .find the prob that it will land inside the circle . Lenght & breadth of the rectangle is 4 &2 m respectivily and diameter of circle is 1/2 m.

Posted by Sanjay Kumar 7 years, 6 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

139 divisible by which number

Posted by Sachin Kumar 7 years, 6 months ago

CBSE > Class 10 > Mathematics

2 answers

Rajiv Ranjan 7 years, 6 months ago

1 ?????

Sachin Kumar 7 years, 6 months ago

I do not know

3Thank You

ANSWER

What is lifeprocess

Posted by Ayush Kumar 7 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Jaya Chaudhary 7 years, 6 months ago

The processes which maintain body functions and are necessary for survival are called life processes. The important life processes are nutrition, transportation, metabolism, reproduction, respiration, and excretion.