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

Sol- let, √5 be rational no. Then, it is in the form of p/q where q not equals to zero and P and q are co-prime. √5=p/q On squaring both side- 5=p²/q² 5q²=p² ____eqation (1) Since, 5 divides p² Therefore, 5 divides p Again, Let, p=5c On squaring both side, P²=25c² 5q² =25c² from equation 1 q² =5c² Since, 5 divides q² Therefore, 5 divides q Since, 5 is the common factor . Therefore, this contradiction Therefore, √5 is an irrational no.

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Shivansh Sharma 6 years, 10 months ago

Yes

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ANSWER

Find the zeros of the quadratic polynomial 5u2+10u and verifiy the relation between the zeros and the coefficient

Posted by Anil Keshri 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Puja Sahoo? 6 years, 10 months ago

U= -2, or 0......

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ANSWER

in what ratio does the x-axis divide the linesegment joining the points (-4,-6)and(-1,7)?find the co-ordinates of the point of division

Posted by Vaitheeswari Eesu 6 years, 10 months ago

CBSE > Class 10 > Mathematics

3 answers

Puja Sahoo? 6 years, 10 months ago

Required ratio is 6:7 and the coordinates of the point of division are (-34/13 , 0)........

1Thank You

Riya Pal 6 years, 10 months ago

Ratio=3:2, Point of divison=-11/5

0Thank You

Satendra Singh 6 years, 10 months ago

The x axis divide the line segment in 6:7 and co-ordinate of point of division are(-34/13,0)

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polynomial x4+7x3+7x2+px+q is exactly divisible by x2+7x+12, then find the value of p and q

Posted by Shalu Singh 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

Factors of x² + 7x + 12 :
x² + 7x + 12 = 0
{tex}\Rightarrow{/tex} x² + 4x + 3x + 12 = 0
{tex}\Rightarrow{/tex} x(x + 4) + 3(x + 4) = 0
{tex}\Rightarrow{/tex}(x + 4) (x + 3) = 0
{tex}\Rightarrow{/tex} x = - 4, -3 ...(i)
Since p(x) = x⁴ + 7x³ + 7x² + px + q
If p(x) is exactly divisible by x²+ 7x + 12, then x = - 4 and x = - 3 are its zeroes. So putting x = - 4 and x = - 3.
p(- 4) = (- 4)⁴ + 7(- 4)³ + 7(- 4)² + p(- 4) + q
but p(- 4) = 0
{tex}\therefore{/tex} 0 = 256 - 448 + 112 - 4p + q
0 = - 4p + q - 80
{tex}\Rightarrow{/tex}4p - q = - 80 ...(i)
and p(-3) = (-3)⁴ + 7(-3)³ + 7(-3)² + p(-3) + q
but p(-3) = 0
{tex}\Rightarrow{/tex}0 = 81-189 + 63 - 3p + q
{tex}\Rightarrow{/tex}0 = -3p + q -45
{tex}\Rightarrow{/tex}3p -q = -45 ..........(ii)

On putting the value of p in eq. (i),we get,
{tex}4(-35) - q = - 80{/tex}
{tex}\Rightarrow{/tex}{tex}-140 - q = - 80{/tex}
{tex}\Rightarrow\ {/tex}{tex}-q = 140 - 80{/tex}
{tex}\Rightarrow\ {/tex} {tex}-q = 60{/tex}
{tex}\therefore{/tex} {tex}q = -60{/tex}
Hence, p = -35, q = -60

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ANSWER

In 5 unit i cant be understand anu thing how it can possible that

Posted by Khushpreet Singh 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Gungun_ Lalwani? 6 years, 10 months ago

Whats ur ques????? Plz write it clearly

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ANSWER

In a right angle triangle pqr ,right angle at p is drawn.A point is m is taken on qr such that pm perpendicular qr. Prove that pm ×pm=qm×mr

Posted by Jay Kumar 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

Given: PQR is a triangle right angled at P and M is a point on QR such that PM {tex} \bot {/tex} QR

To prove PM² = QM.MR
Proof: In right triangle PQR,
QR² = PQ²+ PR²(1).............[By Pythagoras theorem]
In right triangle PMQ,
PQ²= PM²+ MQ² (2)...........[By Pythagoras theorem]
In right triangle PMR,
PR²= PM²+ MR²(3) ............[By Pythagoras theorem]
Using (2) and (3), (1) gives
QR² = (PM²+ MQ²) + (PM² + MR²)
{tex}\Rightarrow {/tex} QR² = 2PM²+ MQ²+ MR²
=(MQ + MR)²= 2PM²+ MQ²+ MR²
{tex}\Rightarrow {/tex} MQ² + MR²+ 2MQ.MR
= 2PM²+ MQ² + MR²
{tex}\Rightarrow {/tex} PM²= QM.MR
A litre.
In {tex}\triangle {/tex}QMP and {tex}\triangle {/tex}PMR,
{tex}\angle{/tex} QMP= {tex}\angle{/tex} PMR......[Each equal to 90^o]
{tex}\angle{/tex}MQP {tex} \sim {/tex} {tex}\angle{/tex}PMR.......AA similarity criterion
{tex}\therefore \frac{{QM}}{{PM}} = \frac{{PM}}{{RM}}{/tex} {tex}\therefore {/tex} corresponding sides of two similar triangles
{tex}\Rightarrow {/tex} PM² = QM.RM
{tex}\Rightarrow {/tex} QM.MR

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In board which book l should study with

Posted by Saloni Gupta 6 years, 10 months ago

CBSE > Class 10 > Mathematics

3 answers

Mehakdeep Kaur 6 years, 10 months ago

Ncert(example question also)

1Thank You

Kajal Singh 6 years, 10 months ago

Ncert, previous year questions, all theorems,. Sample paper

1Thank You

Anshika / U Can Call Me Alien ???? 6 years, 10 months ago

NCERT

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Draw a circle of radius 6 cm. Draw a tangent to this cicle making an angle of 30 degree with a line passing through the centre

Posted by Adhithyan P 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Gungun_ Lalwani? 6 years, 10 months ago

Cant draw here...bro

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Solve for x : 1/a+b+c = 1/a+1/b+1/x

Posted by Sanjana ? 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Anshika / U Can Call Me Alien ???? 6 years, 10 months ago

ON solving the above equation we have a quadratic equation (a+b)x^2+(a+b)^2x+ab(a+b) on solving the above equation by formula -b+root over(b^2-4ac) and -b-root over(b^2-4ac) we got -a and -b as roots.

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ANSWER

What is theorem?

Posted by ☺️ ? 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Puja Sahoo? 6 years, 10 months ago

A, proved statement..........

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ANSWER

Find value of k for which one root of the quadratic equation kx^2 - 14x + 8 =0 is six times the other plz send solution ????

Posted by Shivam Tripathi 6 years, 10 months ago

CBSE > Class 10 > Mathematics

3 answers

Shivam Tripathi 6 years, 10 months ago

Give me some hint to solve it

0Thank You

Nisha Kanchan 6 years, 10 months ago

put 6in X in equation

1Thank You

Puja Sahoo? 6 years, 10 months ago

K=3

4Thank You

ANSWER

(CosecA-sinA) (secA-cosA)=1/tanA+cotA

Posted by Karan Yadav 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Yash Kanchan 6 years, 10 months ago

Cosec-sin×sec-cos =lhs cosec-1/cosec×sec-l/sec=cosec2 -1/cosec ×sec2-1/sec=cot2/cosec×tan2/sec=cos2/sin2/1/sin×sin/cos2/1/cos=sin×cos RHS=1/sin/cos+cos/sin=sin2+cos2/sin×cos=sin×cos LHS=RHS 2is square

3Thank You

ANSWER

(A+b)hole cube formula

Posted by Bhabani Parida 6 years, 10 months ago

CBSE > Class 10 > Mathematics

2 answers

Shivam Tripathi 6 years, 10 months ago

( a - b)(a^2 + b^2 + ab) ,,,,,,,,or,,, ,,,,,, a^3 + b^3 +3a^2b + 3ab^2,,,,,, (^= power)

1Thank You

Purvanshi Yadav 6 years, 10 months ago

(a+b)3= a3+ b3+ 3a2b+ 3ab2

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In the given figure ∆ ABC ~ ∆ DEF find the length of the sides of a ∆ . Given : AB= x-1 AC= 4x BC= 2x+2 DE= 8x DF= 8 EF=3x+9

Posted by Prakriti Tyagi 6 years, 10 months ago

CBSE > Class 10 > Mathematics

4 answers

Madhur Dubey ??? 6 years, 10 months ago

Where is the given figure

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Puja Sahoo? 6 years, 10 months ago

Shayad question adhura hai, plz check it again

2Thank You

Shivam Tripathi 6 years, 10 months ago

Anyone can solve

2Thank You

Prakriti Tyagi 6 years, 10 months ago

How to solve this

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ANSWER

1+1=12

Posted by Rohan Patel 6 years, 10 months ago

CBSE > Class 10 > Mathematics

3 answers

Madhur Dubey ??? 6 years, 10 months ago

Yes,1+1=12 but only because of typing mistake ??????????

0Thank You

Harmanjot Singh 6 years, 10 months ago

1+1+10=12

2Thank You

Harmanjot Singh 6 years, 10 months ago

You want to add 10 in this equation ??????

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ANSWER

If cos =4/5 ,find all other trigonometric ratio of angle

Posted by Nakul Verma 6 years, 10 months ago

CBSE > Class 10 > Mathematics

2 answers

Yogita Ingle 6 years, 10 months ago

sin θ = 3/5
tanθ = 3/4
cosec θ = 1/sinθ = 5/3

Sec θ = 1/cos = 5/4
Cot θ = 1/tan = 4/3

1Thank You

Honey ??? 6 years, 10 months ago

Here p=3 ,b=4 and h=5 now use trigo ratios sin=p/h ,tan=p/b and so on.......

1Thank You

ANSWER

2x+5x=0

Posted by Md Arbaaz Ansari 6 years, 10 months ago

CBSE > Class 10 > Mathematics

2 answers

Yogita Ingle 6 years, 10 months ago

2x + 5 x = 0
7x = 0
x = 0/7
x = 0

1Thank You

Honey ??? 6 years, 10 months ago

2Thank You

ANSWER

Prove pythagorous theorum

Posted by Raja Bhaiya 6 years, 10 months ago

CBSE > Class 10 > Mathematics

2 answers

Anshika / U Can Call Me Alien ???? 6 years, 10 months ago

Its given in NCERT please read it

1Thank You

Honey ??? 6 years, 10 months ago

It is in the book.

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ANSWER

7sin²ø+cos²ø=44 Show that 1/√3tanø Kisi ko pata ho to bata dena Ya (Raghav Garg) fb messenger pe bata dena

Posted by Raghav Garg 6 years, 10 months ago

CBSE > Class 10 > Mathematics

2 answers

Diksha Chawla 6 years, 10 months ago

your question is incomplete

2Thank You

Anshika / U Can Call Me Alien ???? 6 years, 10 months ago

7sin²θ + 3cos²θ = 4 7sin²θ + ౩(1-sin²θ) = 4 7sin²θ+3-3sin²θ = 4 4sin²θ+3 = 4 4sin²θ=1 sin²θ = 1/4 sinθ = +1/2 sinθ = + π/6 Therefore tan θ = + π/6 then we get tan θ = + 1/√3 As the required answer is tan θ = + 1/√3 (here we considered θ =+ π/6

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4sin square 45 degree +tan square 60 degree +cosec square 30 degree

Posted by Chetna Saini 6 years, 10 months ago

CBSE > Class 10 > Mathematics

3 answers

Chetna Saini 6 years, 10 months ago

Please reply sis

0Thank You

Chetna Saini 6 years, 10 months ago

Yes I done it

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Diksha Chawla 6 years, 10 months ago

As sin 45 degree is 1/√2 so sin square 45degree is (1/√2)^2= 1/2 and its 4times become 2.......... Eq(1)

Similarly tan sqare 60 become 3 ..... Eq(2)

And cosec square 30 become 2.... Eq (3)

On adding all these equations as required in question

We get

2+3+2 = 7

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ANSWER

What is the formula of finding the diagonal of a cuboid and of cube...plz help

Posted by Divya Garg 6 years, 10 months ago

CBSE > Class 10 > Mathematics

2 answers

Chirag Mantri 6 years, 10 months ago

Cuboid => √(l^2 + b^2 + h^2)

2Thank You

Yogita Ingle 6 years, 10 months ago

The main diagonal of any cube can be found my multiplying the length of one side by the square root of 3.
Diagonal of cube = √ 3 × side

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AD is an altitude of an equaleteral triangleABC .on AD as base another triangleADE is constructed .prove that area of triangleMDE:area of triangleABCis 3:4

Posted by Sheila Sharma 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Gaurav Seth 6 years, 10 months ago

Given ABC and ADE are equilateral triangles. Let AB=BC=CA = a Recall that the altitude of an equilateral triangle is √3/2 times its side Hence AD = √3a/2ΔABC ~ ΔADE [By AAA similarity criterion] [Since ratio of areas of two similar triangles is equal to the ratio of their corresponding sides]

Hence area(ADE):area(ABC)=3:4.

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ANSWER

If the squared difference of the zeroes of the quadratic equation polynomial F(x)=x2+px+45 is equal to 144,find the value of 'p'.

Posted by Shriyanshi Gupta 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Abhinav Dutta 6 years, 10 months ago

Let the zerors be x and y Sum of zeros= x+y = -p Product of zeroes= xy = 45 A.T.Q ( x-y)2 = 144 x2 + y2 -2xy = 144 We write x2+y2 as (x+y)2 -2xy (x+y)2 -2xy -2xy = 144 (x+y)2 -4xy = 144 On putting above values (-p)2 -4(45) = 144 p2 -180 = 144 p2 = 144+ 180 p2 = 324 p = √324 p = 18

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Ex 6.4 q-7

Posted by Manya Singh 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Sneha Singh 6 years, 10 months ago

Seee on this app ncert solutions u will get the answer

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Find the value of a and b for which the following systems of linear equation has infinite no. of solution, 2x+3y=7,2ax+(a+b)y=28

Posted by Yash Deo 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Asmita Joshi 6 years, 10 months ago

a1/a2=2/2a b1/b2=3/(a+b) c1/c2=7/28 In the case of infinitely many solutions, a1/a2=b1/b2=c1/c2 Therefore, 2/2a=7/28 -> 1/a=1/4 -> a=4 Also, 3/a+b=7/28 where a=4 -> 3/4+b=1/4 -> 12=4+b -> b=12-4 -> b=8

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In triangle PQR right angled at Q , PQ =3 cm and PR =6cm. Determine angle QPR and angle PRQ

Posted by Soumya Tiwari 6 years, 10 months ago

CBSE > Class 10 > Mathematics