CBSE > Class 10 > Mathematics | CBSE Board | Homework Help

Let us take √2 is rational We can take x and y √2 = x/y Suppose x and s have a common factor other than 1. Then you have to divide by the common factor √2 = a/ b, where a and b are coprime. So b √2 = a On sq. Both side 2b ka 2 = a ka 2 2 divide a ka 2. So we take another integer c. a = 2c We get 2b ka 2 = 4c ka 2 b ka 2 = 2c ka 2 That means 2 divide b ka 2 and so 2 divides b. Therefore a and b have at least 2 as a common factor. This contradiction has arisen because of our incorrect assumption that √ 2 is rational. So, we conclude that √2 is irrational.

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Geetanand Yadav/ Rao Sahab??????✌??? 6 years, 10 months ago

See in NCERT

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In which the following rational number can be expressed as a terminating decimal?

Posted by Bobby Tanti 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Geetanand Yadav/ Rao Sahab??????✌??? 6 years, 10 months ago

If one number is perfect square and if it is fraction than it should completely divide each other

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In a parallelogram one angle is 4/5 of adjacent angle.determine the angles of parallelogram

Posted by Anshika Anand 6 years, 10 months ago

CBSE > Class 10 > Mathematics

2 answers

Gaurav Seth 6 years, 10 months ago

et the first angle be x and the second angle be y.
Then x = 4y/5
5x = 4y
5x - 4y = 0 ......(1)
Also x + y = 180° .......(2)
Multiply eq (2) by 5
Then 5x + 5y = 900° ......(3)
(3) - (1) we get
9y = 900°
y = 100°
Put this value in eq (2)
x + 100° = 180°
x = 180° - 100°
x = 80°
Therefore, the angles are 80°, 100°, 80°, 100° because in a parallelogram opposite angles are equal.

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Gaurav Seth 6 years, 10 months ago

Solution:

Let one angle be

Other angle =

But sum of adjacent angles in a parallelogram is

Hence the angles of the parallelogram are

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ANSWER

Note on Function and Polynomials

Posted by Bipin Sharma 6 years, 10 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

2/√x+3/√y=2, 4/√x-9/√y=1

Posted by Vaidant Gupta 6 years, 10 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

Check whether X + 2 whole cube equal x cube + 5 is a quadratic equation

Posted by Rohan Bagri 6 years, 10 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

If cosecA+cotA=p then prove that cosA=p^2-1/p^2+1

Posted by Manoj Panth 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

Given, cosec θ + cot θ = p...(i)

We know that, {tex}cosec^2\theta-cot^2\theta=1{/tex}

{tex}\Rightarrow (cosec\theta+cot\theta)(cosec\theta-cot\theta)=1{/tex}

{tex}\Rightarrow p(cosec\theta-cot\theta)=1{/tex}

{tex}\Rightarrow cosec\theta-cot\theta=\frac 1p{/tex} ....(ii)

Adding i and ii, we get

{tex}2cosec\theta=p+ \frac 1p{/tex}

{tex}cosec\theta=\frac{p^2+1}{2p}{/tex}

{tex}\Rightarrow sin\theta= \frac{1}{cosec\theta}=\frac{2p}{p^2+1}{/tex}

We know that,{tex}cos\theta=\sqrt{1-sin^2\theta}=\sqrt{1- \frac{4p^2}{(p^2+1)^2}}=\sqrt{\frac{p^4+1-2p^2}{(p^2+1)^2}}{/tex}

{tex}cos\theta=\sqrt{\frac{(p^2-1)^2}{(p^2+1)^2}}=\frac{p^2-1}{p^2+1} {/tex}

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Prove that rhombus is not a cyclic quadrilateral

Posted by Sunny Raj 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Ram Kushwah 6 years, 10 months ago

In rhombus ABCD

{tex}\angle A=\angle C{/tex}

And angle A and C both are not equal to 90°

So sum of opposite angles is not equal to 180°

So ABCD is not cyclic.

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ANSWER

Tan 26 Cot64

Posted by Nabeela Hasan 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Yogita Ingle 6 years, 10 months ago

tan 26° /cot 64° = tan26° / cot(90°- 26°) = tan26° /tan26° = 1

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ANSWER

One zero of the polynomial 3x3+ 16x2+15x-18is2/3 find the other zero of polynomial

Posted by Keshav Kunal 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Mansi Srivastava 6 years, 10 months ago

2/3 is a zero of polynomial Then,x=2/3 . 3x=2 . 3x-2=0 is a factor Divide the polynomial by 3x-2 Factorise the remainder and find the zeroes

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if the point x y is equidistant from the points a minus b a + b and minus a minus b a + b prove that X + b is equal to 0

Posted by Nikita Mittal 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Mansi Srivastava 6 years, 10 months ago

Let the point (x,y), (a-b,a+b) & (-a-b,a+b) be A, B & C Respectively.find lenght AB & AC THEN equate them as they are equal.

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how to i learn maths ??

Posted by Harish Rajpurohit 6 years, 10 months ago

CBSE > Class 10 > Mathematics

3 answers

Gautam Prajapati 6 years, 10 months ago

You can't learn maths or anyone cannot learn maths it is not possible....... You have to do practice and hard practice for it.... ☺️☺️

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Naitik Garg 6 years, 10 months ago

We cannot learn math it only wants practise and practise

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Harshit Khandelwal 6 years, 10 months ago

Maths is not be learn by anyone because it want practise and only practise

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ANSWER

use euclid division lemma to show that the square of any positive integeris either the form of 3m or 3mplus 1 for some integer

Posted by Jay Sharma 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Gaurav Seth 6 years, 10 months ago

let ' a' be any positive integer and b = 3.

we know, a = bq + r , 0 < r< b.

now, a = 3q + r , 0<r < 3.

the possibilities of remainder = 0,1 or 2

Case I - a = 3q

a² = 9q²

= 3 x ( 3q²)

= 3m (where m = 3q²)

Case II - a = 3q +1

a²= ( 3q +1 )²

= 9q² + 6q +1

= 3 (3q² +2q ) + 1

= 3m +1 (where m = 3q² + 2q )

Case III - a = 3q + 2

a² = (3q +2 )²

= 9q²+ 12q + 4

= 9q² +12q + 3 + 1

= 3 (3q² + 4q + 1 ) + 1

= 3m + 1 where m = 3q² + 4q + 1)

From all the above cases it is clear that square of any positive integer ( as in this case a²) is either of the form 3m or 3m +1.

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ANSWER

If theta and beta are zeros of the quadratic polynomial (fx) xsquare minus p bracket X + 1) - C show that( theta + 1) bracket beta + bet + 1)equal to 1-c

Posted by Shamaim Ambarin 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

Since, α and β are the zeroes of the quadratic polynomial
f(x) = x² - p (x+1) - c=x²-px-(p+c)
So A=1 B=-p,C=-(p+c)
Sum of the zeroes α + β = {tex}-\frac BA{/tex}=P
Product of the zeroes αβ ={tex}\frac CA{/tex}=- (p + c)
(α + 1)(β + 1)
= αβ + α + β + 1
= αβ + (α + β) + 1
=-(p+c)+p+1
=-p-c-p+1
=1-c
Hence proved

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ANSWER

Ncert exemplar solutions

Posted by Divyansh Singh 5 years, 8 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

If sec theta + tan theta =p. Find the value of sec theta and sin theta

Posted by Ayushi Jain 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Puja Sahoo? 6 years, 10 months ago

Sin theta= p square - 1/ p square +1...... sec thita= p square +1/2p ..........

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An iron pillar has lower part in the form of right circular cylinder and the upper part In the form of right circular cone .the radius of the base of each of the cone and cylinder is 8 cm. The cylindrical part is 240 cm high and the conical part is 36 cm high. find the weight of the pillar , if 1 cm cube of iron weights 7.5g

Posted by Harsh Vardhan Gautam 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Gaurav Seth 6 years, 10 months ago

Let r₁ and r₂ cm be the radii of the base of the cylinder and cone respectively. Then, r₁ = r₂ = 8 cm Let h₁ and h₂ cm be the heights of the cylinder and the cone respectively. Then,

h₁ = 240 cm and h₂ = 36 cm

Now, Volume of the cylinder = r₁²h₁ cm³ = ( x 8 x 8 x 240) cm³ = ( x 64 x 240) cm³
Volume of the cone =
Total volume of the iron = Volume of cylinder + Volume of the cone

Hence, total weight of the pillar = Volume x Weight per cm³
= (22 x 64 x 36 x 7.5) = 380.16 kg

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ANSWER

A circle touches the side BC and triangle abc at P and touches AB and AC produced at Q and R respectively. prove that AQ equal to 1/2 (perimeter of triangle ABC)

Posted by Harsh Vardhan Gautam 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Gaurav Seth 6 years, 10 months ago

Sol:

Given: A circle touching the side BC of ΔABC at P and AB, AC produced at Q and R respectively.

RTP: AP = 1/2 (Perimeter of ΔABC)

Proof: Lengths of tangents drawn from an external point to a circle are equal.
          ⇒ AQ = AR, BQ = BP, CP = CR.
          Perimeter of ΔABC = AB + BC + CA
                                      = AB + (BP + PC) + (AR – CR)
                                  = (AB + BQ) + (PC) + (AQ – PC) [AQ = AR, BQ = BP, CP = CR]
                                      = AQ + AQ
                                = 2AQ

           ⇒ AQ = 1/2 (Perimeter of ΔABC)

           ∴ AQ is the half of the perimeter of ΔABC.

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If p(e)=5×10-2 then find p(e bar)

Posted by Akshay Lal 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Gaurav Seth 6 years, 10 months ago

Given ,p(e)=5×10-2 = 5/100

therefore,P(E)=0.05

P(E) + P(Not E) = 1
⇒ 0.05 + P(Not E) = 1
∴ P(Not E) = 1 – 0.05 = 0.95

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If tanA+sinA=m and tanA - sinA =n, show that m^2−n^2= 4√mn

Posted by Ankul Prajapati 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Ayushi Jain 6 years, 10 months ago

Lhs (tan A+sin A )(tan A-sin A) _(tanA+sin A)(tan A-sin A) =(2tan A)(2sin A) =4tan A sin A Rhs 4√(tanA+sin A)(tanA-sinA) 4√tan Square A-sin square A 4√tan square A_sin square A 4√sin square A/cos square A_sin square A 4√sin square A (1/cos square A-1) 4√sin square A (sec square A-1) 4√sin square A tan square A 4 sin A tan A Hence proved

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If P(e)=5×10-2 then find P(e)

Posted by Akshay Lal 6 years, 10 months ago

CBSE > Class 10 > Mathematics

2 answers

Naitik Garg 6 years, 10 months ago

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Puja Sahoo? 6 years, 10 months ago

48.....??

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Find the coordinate of the points of trisection of the line line segment joining origin O and R(6,-9)

Posted by Ankul Prajapati 6 years, 10 months ago

CBSE > Class 10 > Mathematics

2 answers

Naitik Garg 6 years, 10 months ago

(2,-3) and (4,-6)

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Piyush Raj Sahni 6 years, 10 months ago

(2,-3) and (4,-6)

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Which term of the AP 3,15,27,39.....will be 120 more than its 21st term

Posted by Ankul Prajapati 6 years, 10 months ago

CBSE > Class 10 > Mathematics

2 answers

Chirag Jain 6 years, 10 months ago

a=3 d=15-3 =12 A21 = a+20d =3+20×12 =3+240 =243 An=243+120 =363 An=a+(n-1)d 363=3+(n-1)12 363-3=12n-12 360+12=12n 372=12n 372/12=n 31=n Ans=31

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Raj Kumar 6 years, 10 months ago

31 st term

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two men on either side of a 75 M high building and in line with a base of building observe the angle of elevation of the top of the building is 30 degree and 60 degree find the distance between the two men

Posted by Aashutosh Upadhyay 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Raj Kumar 6 years, 10 months ago

100 root 3

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Find Median of -: CLASS FREQUENCY 5-10 2 10-15 12 15-20 2 20-25 4 25-30 3 30-35 4 35-40 3

Posted by Naveen Lamba 6 years, 10 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

Prove that 3+2 √5 is irrational

Posted by Aashutosh Upadhyay 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Chirag Jain 6 years, 10 months ago

Let us assume 3+√5 is a rational number But all the rational no. are in the form p/q So we can also write 3+√5 in p/q form and when we divide p/q with any of their common factor we get a/b which are coprime 3+√5 = a/b √5 = a/b-3 √5= a-3b/b But we know a-3b/b is a rational no But √5 is an irrational number => Our assumption is wrong. => 3+√5 is an irrational no

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Two men on either side of a 75 m high building and in a line with a base of building observe the angle of elevation of the top of building as 30* and 60*. Find the distance between the two men

Posted by Rishu Jha 6 years, 10 months ago

CBSE > Class 10 > Mathematics