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ANSWER

1÷(x-1)(x-2)+1÷(x-2)(x-3)+1÷(x-3)(x-4)=1/6 solve the value of x

Posted by Rishi Raj 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

We have,
{tex}\frac{1}{{(x - 1)(x - 2)}} + \frac{1}{{(x - 2)(x - 3)}}{/tex} {tex}+ \frac{1}{{(x - 3)(x - 4)}} = \frac{1}{6}{/tex}
{tex}\Rightarrow{/tex}{tex} (x - 3)(x - 4) + (x - 1)(x - 4) + (x - 1)(x - 2) = {/tex}{tex}\frac{1}{6}{/tex}{tex}(x - 1)(x - 2)(x - 3)(x - 4){/tex}
[{tex}\because{/tex} Multiplying both sides by (x -1)(x - 2)(x - 3)(x - 4)]
{tex}\Rightarrow{/tex} {tex}x^2 - 4x - 3x + 12 + x^2 - 4x - x + 4 + x^2 - 2x - x + 2 ={/tex} {tex}\frac{1}{6}{/tex}{tex}[(x - 1)(x - 2)(x - 3)(x - 4)]{/tex}
{tex}\Rightarrow{/tex} {tex}3x^2 - 15x + 18 ={/tex} {tex}\frac{1}{6}{/tex}{tex}(x - 1)(x - 2)(x - 3)(x - 4){/tex}
{tex}\Rightarrow{/tex} {tex}3(x^2 - 5x + 6) ={/tex} {tex}\frac{1}{6}{/tex}{tex}(x - 1)(x - 2)(x - 3)(x - 4){/tex}
{tex}\Rightarrow{/tex} {tex}18[x^2 - 3x - 2x + 6] = (x - 1)(x - 2)(x - 3)(x - 4){/tex}
{tex}\Rightarrow{/tex} {tex}18[x(x - 3) - 2(x - 3)] = (x - 1)(x - 2)(x - 3)(x - 4){/tex}
{tex}\Rightarrow{/tex} {tex}18(x - 3)(x - 2) = (x - 1)(x - 2)(x - 3)(x - 4){/tex}
{tex}\Rightarrow{/tex} 18 = (x - 1)(x - 4)
{tex}\Rightarrow{/tex} 18 = x² - 4x - 1x + 4
{tex}\Rightarrow{/tex} x² - 5x + 4 - 18 = 0
{tex}\Rightarrow{/tex} x² - 5x - 14 = 0
In order to factorize x² - 5x - 14, we have to find two numbers 'a' and 'b' such that.
a + b = - 5 and ab = -14
Clearly, -7 + 2 = -5 and (-7)(2) = -14
{tex}\therefore{/tex} a = -7 and b = 2
Now,
x² - 5x - 14 = 0
{tex}\Rightarrow{/tex} x² - 7x + 2x - 14 = 0
{tex}\Rightarrow{/tex} x(x - 7) + 2(x - 7) = 0
{tex}\Rightarrow{/tex} (x - 7)(x + 2) = 0
{tex}\Rightarrow{/tex} x - 7 = 0 or x + 2 = 0
{tex}\Rightarrow{/tex} x = 7 or x = -2

ANSWER

Draw the graph of the equations 4x-5y+16=0 and 2x +y-6=0 determine the vertices of the triangle formed by these lines and the x axis

Posted by Ishesh Kumar 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

Graph of {tex}4x - 5y + 16 = 0{/tex}
{tex}4x - 5y + 16 = 0{/tex} {tex}\Rightarrow{/tex} {tex}5y = 4x + 16{/tex}
{tex}\Rightarrow \quad y = \frac { ( 4 x + 16 ) } { 5 }{/tex}.......(i)
Table for {tex}4x - 5y +16 = 0.{/tex}

x	-4	1	6
y	0	4	8

Now, plot the points A (-4, 0), B(1, 4) and C(6, 8) on the graph paper.
Join AB and BC to get the graph line ABC. Extend it on both ways.
Thus, the line ABC is the graph of 4x - 5y +16 = 0.

Graph of {tex}2x + y - 6 =0{/tex}
{tex}2x + y - 6 = 0{/tex} {tex}\Rightarrow{/tex} {tex}y = (6 - 2x){/tex}. ... (ii)
Table for {tex}2x + y - 6 = 0.{/tex}

x	0	2	3
y	6	2	0

On the same graph paper as above, plot the points P(0,6), Q(2,2) and R(3,0).
Join PQ and QR to get the graph line PQR. Extend it on both ways.
Thus, the line PQR is the graph of 2x + y - 6 = 0.

The two graph lines ABC and PQR intersect at the point B(1,4). x = 1 and y = 4 is the solution of the given system of equations.
These lines form {tex}\triangle{/tex}BAR with the x-axis, whose vertices are B(l, 4), A (-4, 0) and R(3, 0).
The two graph lines ABC and PQR intersect at the point B(1, 4).
{tex}\therefore{/tex} x = 1 and y = 4 is the solution of the given system of equations.
These lines form {tex}\triangle{/tex}BAR with the x-axis, whose vertices are B(1, 4), A (-4, 0) and R(3, 0).

ANSWER

x¹+x²+x³+x⁴+.........x¹°° is a polynomial or not Give reason

Posted by Aakanksha Goyal 7 years, 2 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

Prive that: tan /(1+ tan^2)^2 +cot/(1+cot^2)^2 =sin×cos

Posted by Vaibhav Bhardwaj 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

LHS {tex}=\frac{\tan A}{\left(1+\tan ^{2} A\right)^{2}}+\frac{\cot A}{\left(1+\cot ^{2} A\right)^{2}}{/tex}
{tex}=\frac{\tan A}{\left(\sec ^{2} A\right)^{2}}+\frac{\cot A}{\left(\ cosec ^{2} A\right)^{2}}{/tex}
{tex}=\frac{\sin A}{\cos A}{/tex} . cos⁴ A {tex}+\frac{\cos A}{\sin A}{/tex} . sin⁴ A
= sin A cos³ A + cos A sin³ A
= cos A sin A (cos² A + sin² A)
= sin A cos A = RHS

ANSWER

A and B are the tangents drawn from an external point to the circle .where O is the center and OP =diameter of circle.Prove that ∆ABP is a equilateral triangle.

Posted by Vaibhav Bhardwaj 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

Join OP, Suppose OP meets the circle at Q. Join AQ.

We have,
OP = diameter
{tex} \Rightarrow{/tex} OQ + PQ = diameter
{tex} \Rightarrow{/tex} PQ = diameter - radius
{tex} \Rightarrow{/tex} PQ = radius.
Therefore, OQ = PQ = radius.
Thus, OP is the hypotenuse of right triangle OAP and Q is the mid-point of OP.
Now,
In {tex} \triangle{/tex}OPA,
{tex} \sin \angle OPA = \frac{{OA}}{{OP}} = \frac{r}{{2r}}{/tex} [Give OP is the diameter of the circle]
{tex} \Rightarrow \sin \angle OPA = \frac{1}{2} = \sin 30^\circ{/tex}
{tex} \Rightarrow \angle OPA = 30^\circ{/tex}
Similarly, it can be proved that {tex} \angle OPB = 30^\circ {/tex}.
Now, {tex} \angle APB = \angle OPA + \angle OPB{/tex} {tex}= 30^\circ + 30^\circ = 60^\circ{/tex}
In {tex} \triangle{/tex}PAB,
PA = PB [Lengths of tangents drawn from an external point to a circle are equal]
{tex}\Rightarrow \angle PAB = \angle PBA{/tex} ....(i) [Equal sides have equal angles opposite to them]
{tex} \angle PAB + \angle PBA + \angle APB = 180^\circ {/tex} [Angle sum property]
{tex}\Rightarrow \angle PAB + \angle PAB = 180^\circ - 60^\circ = 120^\circ {/tex} [Using (i)]
{tex} \Rightarrow 2\angle PAB = 120^\circ{/tex}
{tex} \Rightarrow \angle PAB = 60^\circ{/tex} ....(ii)
From (i) and (ii)
{tex} \angle PAB = \angle PBA = \angle APB = 60^\circ {/tex}
{tex}{/tex} Therefore, {tex} \triangle{/tex}PAB is an equilateral triangle.

ANSWER

Sin 90=

Posted by Piyush Garg 7 years, 2 months ago

CBSE > Class 10 > Mathematics

5 answers

Bidhishree Sahu 7 years, 2 months ago

1 obviously

Subhranshu Mallick 7 years, 2 months ago

1 LOL

Subhadeep Kundu 7 years, 2 months ago

Sharlin Thomas 7 years, 2 months ago

Ncert mai clearly diya hai dude!

Shalu Gupta 7 years, 2 months ago

ANSWER

sin-cos+1÷sin+cos-1=1÷sec-tan

Posted by Kishan Singh 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

We have to prove that,{tex}\frac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} = \frac{1}{{\sec \theta - \tan \theta }}{/tex} using identity {tex}sec^2\theta=1+tan^2\theta{/tex}

LHS = {tex}\frac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} {/tex}{tex} = \frac{{\tan \theta - 1 + \sec \theta }}{{\tan \theta + 1 - \sec \theta }}{/tex} [ dividing the numerator and denominator by {tex}\cos{\theta}{/tex}.]

{tex} = \frac{{(\tan \theta + \sec \theta)-1 }}{{(\tan \theta - \sec \theta )+1}}{/tex}{tex}=\frac{\{{(\tan\theta+\sec\theta)-1\}}(tan\theta-\sec\theta)}{\{{(\tan\theta-\sec\theta)+1\}}(\tan\theta-\sec\theta)}{/tex} [ Multiplying and dividing by {tex}(\tan{\theta}-\sec{\theta}){/tex}]

{tex}=\frac{{(\tan^2\theta-\sec^2\theta)-}(tan\theta-\sec\theta)}{\{{(\tan\theta-\sec\theta)+1\}}(\tan\theta-\sec\theta)}{/tex} [{tex}\because (a-b)(a+b)=a^2-b^2{/tex}]

{tex} = \frac{{-1-\tan \theta + \sec \theta }}{{(\tan \theta - \sec \theta+1)(\tan{\theta}-\sec{\theta}) }}{/tex}[{tex}\because \tan^2\theta-\sec^2\theta=-1{/tex}]

{tex}=\frac{-(\tan\theta-\sec\theta+1)}{(\tan\theta-\sec\theta+1)(\tan\theta-\sec\theta)}{/tex}{tex}=\frac{-1}{\tan{\theta}-\sec{\theta}}{/tex}

{tex} = \frac{1}{{\sec \theta - \tan \theta }}{/tex}=RHS

Hence Proved.

ANSWER

What is tangent and circumferance in circle?

Posted by Saurabh Bisht 7 years, 2 months ago

CBSE > Class 10 > Mathematics

1 answers

Shobhit Jaryal 7 years, 2 months ago

Tangent is line draw outside of circle and get in contact with circle on a certain one point... Circumferance of circle is 2πr

ANSWER

the sum of radius of the base and the height of a cylinder is 37 meter if the total surface area of cylinder be 1628square meter find the volume

Posted by Vishal Verma 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

Let the radius of the base and the height of the solid cylinder be r m and h m respectively.
Then r + h = 37 . . . (1) . . .[Given]
Total surface area of the solid cylinder = 1628 m²
⇒ 2{tex}\pi{/tex}r(h + r) = 1628
⇒ 2{tex}\pi{/tex}r(37) = 1628 . . . [Using (1)]
⇒ 2 × {tex}22\over7{/tex} × r × 37 = 1628
⇒ r = {tex}{{1628}\times7}\over2\times22\times37{/tex}
⇒ r = 7 m
Putting the value of r in (1), we get
7 + h = 37
⇒ h = 37 – 7 = 30 m
∴ Circumference of the base of the solid cylinder = 2{tex}\pi{/tex}r
= 2 ×{tex}22\over7{/tex}× 7 = 44 m
and volume of the cylinder = {tex}\pi{/tex}r²h
={tex}22\over7{/tex}× (7)² × 30 = 4620 m³

ANSWER

A Ladder 15 metres long just reaches the top of a vertical wall if the ladder makes an angle of 60° with the wall find the height of the wall

Posted by Shyam Agrawal 7 years, 2 months ago

CBSE > Class 10 > Mathematics

2 answers

Yo Yo Honey 7 years, 2 months ago

15/2

Ayush Chauhan 7 years, 2 months ago

15√3/2

ANSWER

AP 3,8,13-----253

Posted by Ritik Thakur 7 years, 2 months ago

CBSE > Class 10 > Mathematics

1 answers

Kannu Kranti Yadav 7 years, 2 months ago

What we have to find here

ANSWER

From point A on a Straight road the angle of alevationof the top of a vertical pole tower situated on the roof of the vertical building on the same road is theta.The angle of the elevation of the bottom of the tower from a point B on the road is again theta .The height of the building is 50m . If AB:BY is 2:5,where y is the base of building, then showthat the height of the tower is 20m

Posted by Rajpara Jay Prakashbhai 7 years, 2 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

How to find the zeroes of cubic polynomial Actually I want exact procedure ...pls help

Posted by Rachit Dwivedi 7 years, 2 months ago

CBSE > Class 10 > Mathematics

2 answers

Rachit Dwivedi 7 years, 2 months ago

Acccha ....zyada gyan mat baato jitna poocha h utna batao agar pta h to

Rajpara Jay Prakashbhai 5 years, 8 months ago

Itna bhi nahi aaata hamare esh ki kab pragati hogi gawaaar

ANSWER

What is the value of alpha minus beta

Posted by Harsh Matheria 7 years, 2 months ago

CBSE > Class 10 > Mathematics

1 answers

Subhadeep Kundu 7 years, 2 months ago

√b²-4ac/a

ANSWER

Write the quadratic polynomial whose roots are 2 and -3

Posted by Aakanksha Dhumal 7 years, 2 months ago

CBSE > Class 10 > Mathematics

1 answers

Kannu Kranti Yadav 7 years, 2 months ago

Roots of polynomial are 2 and -3 . So we get the sum of roots = -1 and product of roots =-6. So polynomial is x²+1x-6.

ANSWER

The sum of the 4th and 8th terms of an AP is 24 ,and the sum of the 6th and the 10th term is 34. Find the 4th term of AP.

Posted by Piyush Aggarwal 7 years, 2 months ago

CBSE > Class 10 > Mathematics

2 answers

Piyush Aggarwal 7 years, 2 months ago

How

Vinayak Avasthi 7 years, 2 months ago

Fourth term is 7

ANSWER

Find the value of tan A, where A is an acute angle and cosec A= 2/√3

Posted by Gaurav Dwivedi 5 years, 8 months ago

CBSE > Class 10 > Mathematics

1 answers

Utkarsh Singh 7 years, 2 months ago

Under root 3

ANSWER

Prove that ac +Ab + bc greater 2ap + pq

Posted by Adu Choudhry 7 years, 2 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

Sin (A+B)=?

Posted by Abhishek Mishra 7 years, 2 months ago

CBSE > Class 10 > Mathematics

1 answers

Sagar Jogi 7 years, 2 months ago

Let A =B=30 Sin(A+B) =sin(30 +30)=sin60=√3/2

ANSWER

Sides of the similar triangle 4cm and 9cm are in the ratio of?

Posted by Harshitha Harshi 7 years, 2 months ago

CBSE > Class 10 > Mathematics

1 answers

Harneet Kaur Arora 7 years, 2 months ago

2/3

ANSWER

Find the value of -8-(10) using no. Line

Posted by Abhinav Singh Rana 7 years, 2 months ago

CBSE > Class 10 > Mathematics

1 answers

Yogita Ingle 7 years, 2 months ago

- 18

ANSWER

Can anybody tell me that what is da difference between a tangent to a circle and a secant

Posted by Punit Khandelwal 7 years, 2 months ago

CBSE > Class 10 > Mathematics

3 answers

Shagun Chaudhary 7 years, 2 months ago

Tanget touch at a single point whereas secant touch at two points....

Amit Gupta 7 years, 2 months ago

Tangent touches only one point of the perimeter of cirle (eg football on the ground ) and secant touches two points of the perimeter of circle....

Ak Ak 7 years, 2 months ago

Tangent touch at only one point but secant at two point

ANSWER

2(2)

Posted by Vikash Kumar 7 years, 2 months ago

CBSE > Class 10 > Mathematics

2 answers

Aashu Kumar 7 years, 2 months ago

Utkarsh Pandey 7 years, 2 months ago

ANSWER

Find three consecutive terms which are in AP whose sum is 24 and product is 440

Posted by Puttaraj Javalagi 7 years, 2 months ago

CBSE > Class 10 > Mathematics

2 answers

Aashu Kumar 7 years, 2 months ago

(a+d) +a+(a+d) =24 and(a-d)(a) (a+d) =440

Shalu Gupta 7 years, 2 months ago

(a-d)+a+(a+d) =24 and (a-d)(a)(a+d)=440

ANSWER

What is the formula of sin

Posted by Pawan Dwivedi 7 years, 2 months ago

CBSE > Class 10 > Mathematics

4 answers

Aashu Kumar 7 years, 2 months ago

Perpendicular /hypotenuse

Sagar Jogi 7 years, 2 months ago

SinA=tanA.cosA

Harsh Naik 7 years, 2 months ago

P/H

Ananya Pandey 7 years, 2 months ago

Perpendicular/hypotenuse

ANSWER

Math ka sllyebus

Posted by Raju Nigam 7 years, 2 months ago

CBSE > Class 10 > Mathematics

1 answers

#Abhisht Singh 7 years, 2 months ago

Chapter 1st to chapter 9th

ANSWER

2:3

Posted by Genius ???? 5 years, 8 months ago

CBSE > Class 10 > Mathematics

3 answers

Aashu Kumar 7 years, 2 months ago

2/3

Sagar Jogi 7 years, 2 months ago

It means 2/3

Dheeraj Verma 7 years, 2 months ago

What is the question ? Clarify your question .

ANSWER

Prove that 6-root3 is irrational.

Posted by Harshitha Harshi 7 years, 2 months ago

CBSE > Class 10 > Mathematics

2 answers

Aashu Kumar 7 years, 2 months ago

Yes

Kannu Kranti Yadav 7 years, 2 months ago

Let 6√3 be a rational number , so it can be written in the form of a/b where b≠0 and a and b are co prime numbers ......... .......... 6√3=a/b .. now √3 = a/b-6 , as we know that √3 is an irrational number , therefore our assumption is wrong. Hence due to contradiction 6√3 is an irrational number .hence proved .