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

By using paper , pencil , scale ,compass ,eraser ,and never forget to use sharped pencil as it would deduct your marks Also never use pen in condtruction. Try to write the dimensions of every angle and line as it gives a good presentation. Also write the step of construction.

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How to draw the graph of zeros of a polynomial

Posted by Kunal Choudhary 6 years, 11 months ago

CBSE > Class 10 > Mathematics

1 answers

Jashan Singh 6 years, 11 months ago

First of find zeros and Make the table like Ex. X when put the value of X then the and. Ans. Is y

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Find lcm of 120 and140 by using fundamental theorm of arithmetic

Posted by Aryan Thind 6 years, 11 months ago

CBSE > Class 10 > Mathematics

1 answers

Puja Sahoo? 6 years, 11 months ago

144=2*2*2*2*3*3 120=2*2*2*3*5 lcm=2*2*2*2*3*3*5=720 hcf=2*2*2*3=24

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(cos^2 20+cos^2 70 divide by sin^2 20+sin^2 70)+sin 64 +cos64×sin20

Posted by Aryan Thind 6 years, 11 months ago

CBSE > Class 10 > Mathematics

2 answers

Puja Sahoo? 6 years, 11 months ago

Ceck yur question, i think it should be, cos64×sin26 at last..........

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Aryan Thind 6 years, 11 months ago

Plz give answer fast

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1.73 which angle value of tan

Posted by Vishal Kumar 6 years, 11 months ago

CBSE > Class 10 > Mathematics

3 answers

Puja Sahoo? 6 years, 11 months ago

Yesss, its Tan 60°.....

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Neha Sirur 6 years, 11 months ago

'Tan 60' as tan 60 stands fr √3 and value of it is 1.73

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Lavina Rajput ??? 6 years, 11 months ago

Complete ur question!

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Divide 24 in 3 parts such that they are in AP and their product is 440 ??

Posted by Anshika & Aliens???? 6 years, 11 months ago

CBSE > Class 10 > Mathematics

1 answers

Lavina Rajput ??? 6 years, 11 months ago

Solution:- Let the required terms of the given AP be a-d, a and a+d * Where the first term is a-d* The common difference = d* Given : The sum of the three parts = 24* ∴ (a-d)+(a)+(a+d) = 24 3a = 24 # a = 8# Given : The product of these three terms = 440 ∴ (a-d) (a) (a+d) = 440# (8-d) (8) (8+d) = 440# - 8d² + 512 = 440# - 8d² = 440 - 512# - 8d² = - 72# d² = 72/8# d² = 9# d = √9# d= 3# So the three required terms of AP is 8 - 3 = 5 ; 8 and 8 + 3 = 11 Three terms are 5, 8, 11

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In an ap if the common different -4,and the seventh term is 4, then find the first term

Posted by Divyanshu Rawat 6 years, 11 months ago

CBSE > Class 10 > Mathematics

6 answers

Augustya Singh 6 years, 11 months ago

Given d=-4 and T7=4 a+ 6d=4 a+6(-4)=4 a=4+24 a=28

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Purvanshi Yadav 6 years, 11 months ago

Given that a7=4 a+6d=4 a+6(-4)=4 (d=-4) a-24=4 a=24+4 a=28

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Aryan Thind 6 years, 11 months ago

A=12

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Diksha Rana 6 years, 11 months ago

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Diksha Rana 6 years, 11 months ago

_12

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Anshika & Aliens???? 6 years, 11 months ago

A=28

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If tanA+cosA=p,than Find value of cosecA.

Posted by Rekhansh Chaturvedi 6 years, 11 months ago

CBSE > Class 10 > Mathematics

3 answers

Poonam Bais 6 years, 11 months ago

Question me tan A +cosA=p h yrr

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Poonam Bais 6 years, 11 months ago

And again plz

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Nikita Sharma 6 years, 11 months ago

I have already answered it earlier

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A train travels a distance of 480km at a uniform speed. If the speed had been 8km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train

Posted by Akanksha Priya 6 years, 11 months ago

CBSE > Class 10 > Mathematics

1 answers

Anjali Kumari? 6 years, 11 months ago

40 km/h

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If 1+cos theta /1-cos theta =16/9 ,find 1+cot theta /1-cot theta

Posted by Poonam Bais 6 years, 11 months ago

CBSE > Class 10 > Mathematics

0 answers

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Prove that CosA-SinA+1/CosA+SinA-1 =1/CosecA+CotA

Posted by Premangshu Das 6 years, 11 months ago

CBSE > Class 10 > Mathematics

0 answers

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to prove root 2is irrational

Posted by Mandeep Singh 6 years, 11 months ago

CBSE > Class 10 > Mathematics

1 answers

Amna Nasim 6 years, 11 months ago

let us assume that root 2 is rational no. which is in the form of p/q , where q is not equal to 0. therefore, root 2 = p/q squaring both the sides 2 = p^2/q^2 2q^2 = p^2 ....(1) Here, 2 is divisible by p^2 therefore, 2 is also divisible by p let p = 2c [for some integer c] now, substituting p = 2c in equation (1). therefore, 2q^2 = 4c^2 that is q^2 = 2c^2 this means that 2 is divisible q^2 and so 2 is also divisible by q since p and q have 2 as a common factor this contradict the fact that p & q have no common factor other than 1. But this contradiction has arisen due to our wrong assumption that is root 2 is rational no. so, we conclude that root 2 is irrational.

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CosA/2-sinA=1+sinA/cosA

Posted by Deveshi Nema 6 years, 11 months ago

CBSE > Class 10 > Mathematics

1 answers

Chiranjeevi Kt 6 years, 11 months ago

/ means what ? Then the answer comes

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Find all zeroes of the polynomial if two zeros are (2+root3) and (2-root 3)

Posted by Sunil Sood 6 years, 11 months ago

CBSE > Class 10 > Mathematics

2 answers

Puja Sahoo? 6 years, 11 months ago

Bt, polynomial hai kahan...?????

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Chetna Pandey ? 6 years, 11 months ago

Where is the polynomial

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Solve b2x/a - a2y/b = ab(a + b) b2x - a2y = 2a2b2

Posted by Vinuthna Hasthi 6 years, 11 months ago

CBSE > Class 10 > Mathematics

1 answers

Aditya Sanga 5 years, 4 months ago

hbj

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How can be 2+2 will equal to 5.

Posted by Harshita Patel 6 years, 11 months ago

CBSE > Class 10 > Mathematics

3 answers

Farhat Shaikh Shaikh 6 years, 11 months ago

Ram

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Farhat Shaikh Shaikh 6 years, 11 months ago

I appreciate you

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Ram Kushwah 6 years, 11 months ago

Mr Yuvraj ,

This is useless thing 2+2=5 is impossible,behave like a good student and do not waste your time.

May be somebody prove this by using 0\0=1,which is again wrong and useless.

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If the areas of two similiar triangles are equql, prove that they are congruent

Posted by Dinesh Somani 6 years, 11 months ago

CBSE > Class 10 > Mathematics

1 answers

Gaurav Seth 6 years, 11 months ago

Q.If the areas of two similar triangles are equal, prove that they are congruent.

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Solution:

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If alpha and beta are the zeros of the quadratic polynomial f x is equal to x squared plus x minus 2 then find the polynomial whose zeros are 2 alpha + 1 and 2 Beta + 1

Posted by Ashwani Kumar 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

It is given that {tex} \alpha{/tex} and {tex} \beta{/tex} are the zeros of the polynomial {tex}f(x)=x^2-x-2{/tex}
Let,{tex}g(x)=k(x^2-Sx+P){/tex} be the required polynomial where 'S' is sum of zeroes and 'P' is product of zeroes.
Sum of zeroes of required polynomial (S)={tex}(2 \alpha+1)+(2 \beta+1)=2( \alpha+ \beta)+2=2 \times1+2=4{/tex}
Product of zeroes of required polynomial (P) = {tex}(2 \alpha+1) \times(2 \beta+1) = 4 \alpha \beta + 2 \alpha + 2 \beta + 1 = 4 \alpha \beta + 2 ( \alpha + \beta ) + 1{/tex}
{tex}=4 \times-2+2 \times1+1=-8+2+1=-5{/tex}
Hence, required polynomial g(x) is {tex}k(x^2-4x-5){/tex}, where k is any non-zero constant.

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X² +100x-300000=0 Solve for x

Posted by Samriddhi Shrivastava 6 years, 11 months ago

CBSE > Class 10 > Mathematics

3 answers

Anushka ? 6 years, 11 months ago

X is equal to - 600 or 500

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Gaurav Seth 6 years, 11 months ago

We see that it is a quadratic polynomial.
so, splitting the middle term,
x²+100x-300000
x²+600x - 500x - 300000 =0
x(x+600) -500(x+600) = 0
(x-500)(x+600) = 0

X = 500 or X = -600

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Yogita Ingle 6 years, 11 months ago

x² + 100x - 300000 = 0
x² + 600x - 500x - 300000 = 0
x(x + 600) - 500 (x + 600) = 0
(x + 600) (x - 500) = 0
x = -600 and x = 500

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3/x+1_1/2 =2/3x_1=5/x+1

Posted by Alok Panwar 6 years, 11 months ago

CBSE > Class 10 > Mathematics

0 answers

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Sample papper 12 answer

Posted by Virendra Singh Rajawat 6 years, 11 months ago

CBSE > Class 10 > Mathematics

0 answers

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The centre of a circle is(2a-1,7) and it passes through the point (-3,-1).if the diameter of the circle is 20 units.the find the value(s) of a

Posted by Satyam Raj Raj 6 years, 11 months ago

CBSE > Class 10 > Mathematics

2 answers

Ram Kushwah 6 years, 11 months ago

Let O(2a-1,7) is the center and A(-3,-1) is on the circumference then

OA=r=20/2=10

{tex}\begin{array}{l}OA^2=10^2=100\\or\;\;(2a-1+3)^2+(7+1)^2=100\\\;\;(2a+2)^2+64=100\\4a^2+8a+4+64=10\\4a^2+8a-32=0\\a^2+2a-8=0\\(a+4)(a-2)=0\\Hene\;a=-4\;and\;2\end{array}{/tex}

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Gawsic Varun 6 years, 11 months ago

I dont know

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APB and CQD are semicircles of diameter 7cm each while ARC and BSD are semicircles of diameter 14cm each find the perimeter of shaded region

Posted by X Y 6 years, 11 months ago

CBSE > Class 10 > Mathematics

1 answers

Amit Saroha 6 years, 11 months ago

Give fig. Also

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If x=3 is one root of the quadratic eq x2-2kx-6=0

Posted by Sonakshi Goel 5 years, 8 months ago

CBSE > Class 10 > Mathematics

7 answers

Augustya Singh 6 years, 11 months ago

x2-2kx-6=0 (3)square-6k-6 9-6k-6=0 3-6k=0 -6k=-3 k=-3/-6 k =1/2

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Murari Barnwal 6 years, 11 months ago

(x)2 -2kx -6=0 ,(3)2-2k(3)-6=0 9-6k-6=0 3-6k=0 So, k=1/2

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Yogita Ingle 6 years, 11 months ago

k = 1/2

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Yogita Ingle 6 years, 11 months ago

sorry wrong solution added.
Correct solution is
x2 - 2kx - 6 = 0
x = 3
(3)2 - 2k(3) - 6 = 0
9 - 6k - 6 = 0
- 6k + 3 = 0
6k = 3
k = 3/6
k = 1/3

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Yogita Ingle 6 years, 11 months ago

x2 - 2kx - 6 = 0
x = 3
(3)2 - 2k(3) - 6 = 0
9 - 6k - 6 = 0
2 - 6k = 0
6k = 2
k = 2/6
k = 1/3

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Jiya Arora????? 6 years, 11 months ago

Let p(x)=x^2-2kx-6 Let X = 3 in p(x) p(3)=(3)^2 - 2k(3) -6 =9-6k-6 =3-6k -3=-6k 3=6k 3/6=k 1/2=k

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X Y 6 years, 11 months ago

Value of k=6

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1-sin A ÷1+sinA = (secA+tanA)²

Posted by Spark Light 6 years, 11 months ago

CBSE > Class 10 > Mathematics

2 answers

Shreyanka Das 6 years, 11 months ago

Very nice explain gaurav seth

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Gaurav Seth 6 years, 11 months ago

1-sinA/1+sinA=(secA-tanA)²

L.H.S

1-sinA/1+sinA

Rationalising the denominator

1-sinA (1-SinA) /1+sinA(1-SinA)

(1-SinA)²/ 1² -(Sin²A)

(1-SinA)²/ 1 -(Sin²A)

(1-SinA)² /Cos² A

[ 1 -Sin²A = cos²A]

(1-SinA/CosA)²

(1/CosA-SinA/CosA)²

(SecA-tanA)²

L.H S = R H.S

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Prove that in a right angel tringal squar of the hypotenuse is equal to sum of the squar of other side

Posted by Sweta Nishad 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

Given : {tex}\triangle \mathrm{ABC}{/tex} is right angle at B
To prove: {tex}A C^{2}=A B^{2}+B C^{2}{/tex}

Construction: Draw {tex}B D \perp A C{/tex}

Proof:In {tex}\triangle{/tex}ABC and {tex}\triangle{/tex}ABD

{tex}\angle{/tex}ABC={tex}\angle{/tex}ADB=90

{tex}\angle{/tex}A is common

{tex}\angle{/tex}ABC{tex}\sim{/tex}{tex}\angle{/tex}ADB
{tex}\Rightarrow \frac{A D}{A B}=\frac{A B}{A C}{/tex}
{tex}\Rightarrow{/tex} AD. AC = AB² ...(i)
Similarly {tex}\Delta \mathrm{BDC} \sim \Delta \mathrm{ABC}{/tex}
{tex}\Rightarrow \frac{C D}{B C}=\frac{B C}{A C}{/tex}
{tex}\Rightarrow{/tex} CD.AC = BC² ...(ii)
Adding (i) and (ii)
AD.AC + CD. AC= AB² + BC²
AC(AD + CD) = AB² + BC²
AC {tex}\times{/tex} AC = AB² + BC²
AC² = AB² + BC²
Hence Proved

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Find the value of .tan²30+3cos²30

Posted by Technical Shamim 6 years, 11 months ago

CBSE > Class 10 > Mathematics

4 answers

Augustya Singh 6 years, 11 months ago

1/3+ 9/4=4+27/12=31/12

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Shreyanka Das 6 years, 11 months ago

Tan²30°=√1/3 Cos²30°=√3/2 So, (√1/3)²+3(√3/2)² 1/9+3x3/4 1/9+9/4 4+81/36 85/36 is the answer

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Bhavya Choubey 6 years, 11 months ago

Sorry answer is 11/12......by mistAke wrong answer posted

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Bhavya Choubey 6 years, 11 months ago

1/3 + 3/2=5/6

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FROM NCERT 8.4 ,5 question and part i

Posted by Akash Kumar 5 years, 8 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

If cos A = 2/5, find the value of 4 + 4 tan2 A.

Posted by Sahil Bhati 6 years, 11 months ago

CBSE > Class 10 > Mathematics

3 answers

Shreyanka Das 6 years, 11 months ago

CosA =2/5=b/h P= √h²-b² =√5²-2² =√25-4=√21 Now, tan²A=(√21)²/2² =21/4 According to question, 4+4×21/4=4+21=25 So the right answer is 25.... Hope it helps you a lot....

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Madhu Singh 6 years, 11 months ago

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Anjali Kumari? 6 years, 11 months ago

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