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

Inner Circumference = 352m Inner Radius = 56m Outer Circumference = 396m Outer Radius = 63m Width = R - r 63m - 56m = 7m Area of track = pie (63-56)(63+56) 22/7 (7×119) 22/7 × 7 × 119 2618

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If 3cos theta-4sin theta =2cos theta+sin theta? Find tan theta

Posted by Bhartendu Vibhu 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

{tex} \Rightarrow 3\cos \theta - 4\sin \theta = 2\cos \theta + \sin \theta {/tex}
{tex} \Rightarrow - 4\sin \theta - \sin \theta = 2\cos \theta -3cos \theta {/tex}
{tex} \Rightarrow - 5\sin \theta = - \cos \theta {/tex}
{tex} \Rightarrow 5\sin \theta = \cos \theta {/tex}
{tex} \Rightarrow \frac{{\sin \theta }}{{\cos \theta }} = \frac{1}{5}{/tex}
{tex} \Rightarrow \tan \theta = \frac{1}{5}{/tex}

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Find the value of k for which the system of equation have a unique solution x-xy=2 ,3x-3y=7

Posted by Abhishek Sharma 7 years, 6 months ago

CBSE > Class 10 > Mathematics

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Prove that 3+5√2 is irrational.

Posted by Account Deleted 7 years, 6 months ago

CBSE > Class 10 > Mathematics

2 answers

Nishi K 7 years, 6 months ago

Let us assume that 3+5√2is rational no. written in the form of p/q where p,q are integer. Suppose p,q have no any common factor . 3+2√5=p/q 5√2=p/q-3 √2 =1/5(p/q-3) Since, 1/5(p/q-3)is rational So, √2 is also a rational . This is contradicts the fact that √2 is irrational. Hence, 3+2√5 is irrational.

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Shubham Kumar 7 years, 6 months ago

Let 3+5√2 is a rational number Then,3+5√2=p/q( rational number always in the form of p/q where q not equal to '0') 3+5√2=p/q 5√2=p/q-3 √2=p/q-3/5 But we know that √2 is an irrational no° . So , 3+5√2 is an irrational number.

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What is zero polynomial

Posted by Ansh Chaudhary 7 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Nishi K 7 years, 6 months ago

A real number k is called a zero of the polynomial.p(x) ,if p(k).

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Solve 35x+23y=209 ; 23x+35y=197

Posted by Lavanya Sampath 7 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Account Deleted 7 years, 6 months ago

Solution can be given by cross multiplication method. X=4 and y=3

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If cosA = 4/5,cosB=12/13,3π/2<A,2<2π,the values of the cos(A+B) is

Posted by Ansh Sharma 7 years, 6 months ago

CBSE > Class 10 > Mathematics

0 answers

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3 root 5 minus 2 root 3

Posted by Priya Dharshini 7 years, 6 months ago

CBSE > Class 10 > Mathematics

0 answers

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6 men and 10 women can finish making pots in 8 days ,while the 4 men and 6 women can finish it in 12 days . Find the time taken by the one man alone from that of one woman alone to finish the work

Posted by Anooja Johnson 7 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Account Deleted 7 years, 6 months ago

The solution is given by cross multiplication method .x=18 and y=-10

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Factorise x-1/x=3

Posted by Ankit Kumar 7 years, 6 months ago

CBSE > Class 10 > Mathematics

0 answers

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Find ten rational number between -1/3 and 1/2

Posted by Bharat Rajpal 7 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Shubham Kumar 7 years, 6 months ago

Multiply both no° by 10 Then we get (-1×10)/(3×10) and (1×10)/2×10) =-10/30 and 10/20 Therefore rational no° are-10/30,-11/30............10/20

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Weitage fir every chapter

Posted by Neha Rani 7 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Jagroop Randhawa 7 years, 6 months ago

11?

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If 3 and -5 are the zeroes of a polynomial 6x CUBE + 7y CUBE - ky -3x-4 find k

Posted by Varsha Sahu 7 years, 6 months ago

CBSE > Class 10 > Mathematics

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2(ax-by)+a+4b=0 2(bx+ay)+b-4a=0 Find the solution of given system of equation by cross multiplication method

Posted by Eshika Pawar 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

The given system of equation may be written as
2(ax - by) + a + 4b = 0
So, 2ax - 2by + a + 4b ..............(i)
2(bx + ay) + b - 4a = 0
so, 2bx+2ay+b-4a=0................(ii)
compare (i) and (ii) with standard form, we get
a₁ = 2a, b₁ = -2b, c₁ = a + 4b
a₂ = 2b, b₂ = 2a, c₂ = b - 4a
By cross multiplication method
{tex} \frac{x}{{ - 2{b^2} + 8ab - 2{a^2} - 8ab}}{/tex} {tex}= \frac{{ - y}}{{2ab - 8{a^2} - 2ab - 8{b^2}}}{/tex} {tex} = \frac{1}{{4{a^2} + 4{b^2}}}{/tex}
{tex} \frac{x}{{ - 2{b^2} - 2{a^2}}} = \frac{{ - y}}{{ - 8{a^2} - 8{b^2}}} = \frac{1}{{4{a^2} + 4{b^2}}}{/tex}
Now, {tex}\frac{x}{{ - 2{b^2} - 2{a^2}}} = \frac{1}{{4{a^2} + 4{b^2}}} {/tex}
{tex}⇒ x = \frac{{ - 1}}{2}{/tex}
And, {tex}\frac{{ - y}}{{ - 8{a^2} - 8{b^2}}} = \frac{1}{{4{a^2} + 4{b^2}}} {/tex}
{tex}⇒ y = 2{/tex}
Therefore, the solution of the given pair of equations are {tex}\frac{{ - 1}}{2}{/tex} and 2 respectively.

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Given triangle ABC Given triangle ABC congruent triangle pqr if a b upon PQ equal to 1 upon 3 then find area triangle of ABC upon area Triangle pqr

Posted by Manjot Jaj 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

{tex}\frac{\text { ar } \triangle \mathrm{ABC}}{\text { ar } \triangle \mathrm{PQR}}{/tex} = {tex}\frac{\mathrm{AB}^{2}}{\mathrm{PQ}^{2}}{/tex} = {tex}\left(\frac{1}{3}\right)^{2}{/tex} = {tex}\frac{1}{9}{/tex}

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Discuss the nature of the roots of the equation.4x²-2x+1=0.

Posted by Kannu Kranti Yadav 7 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Rachana Tewari 7 years, 6 months ago

D=-12

so roots are not real.

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2x^2+x-528 Solve by factorisation

Posted by Tisha Jain 5 years, 8 months ago

CBSE > Class 10 > Mathematics

1 answers

Shubham Kumar 7 years, 6 months ago

By factorization . Values of x are = -33/2 and 32 . GOT it Miss Tara jain

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A quadratic polynomial whose zeroes are -3 and 4

Posted by Pardeep Jatt 7 years, 6 months ago

CBSE > Class 10 > Mathematics

4 answers

Kannu Kranti Yadav 7 years, 6 months ago

Formula for making polynomial=x²-(a+B)x+aB.puting values. x²-(-3+4)x+(-3)(4)=0. x²-1x-12=0, it's the answer.

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Rachana Tewari 7 years, 6 months ago

here sum of the roots, S=-3+4=1

Product of the roots P =(-3)(4)=-12

We know that x²+ (sum of the roots)x + product of roots =0

x²+ x- 12=0

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Shubham Kumar 7 years, 6 months ago

x2 - x - 12 It is the correct answer.

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Shubham Kumar 7 years, 6 months ago

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0.4x + 0.3y =1.7 0.7x -0.2y =0.8 Solve the following system of equation

Posted by Sakina Saify 7 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Jigyansh Dara 7 years, 6 months ago

O.4x+0.3y=1.7 4/10x+ 3/10y=17/10 4x+3y/10=17/10 4x+3y=17 -----(1) Similarly 7x-2y=8-----(2) 4x=17-3y X=17-3y/4---(3) Puting (3) in (2) 7 (17-3y/4)-2y =8 119-21y/4 -2y=8 Taking lcm 119-21y-8y=8×4 -29y=36-119 -29y=-83 Y=3 7x-2 (3)=8 7x=8+6 X=14/2 X=2,y=3

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Find the product of zeroes of -2x^2+kx+6

Posted by Jitender Sahoo 7 years, 6 months ago

CBSE > Class 10 > Mathematics

2 answers

Rachana Tewari 7 years, 6 months ago

Product of zeros=c/a

=6/-2=-3

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Priyanshu Kumar 7 years, 6 months ago

-3

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Find value of p and q if the polynomial x^4+px+2x^2-3x+q is divisible by (x-1)(x+1)

Posted by Tanvi Jindal 7 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Ajit Yadav 7 years, 6 months ago

Np12

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Trignomatry formulas

Posted by Royal Rajput 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

Get formulae from notes : <a href="https://mycbseguide.com/cbse-revision-notes.html">https://mycbseguide.com/cbse-revision-notes.html</a>

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If alpha and beta are the zeros of polynomial f(x)=x2+x-2,find the value of 1/alpha-q/beta

Posted by Shubh Agrawal 7 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Shubh Agrawal 7 years, 6 months ago

Answer=-3/2

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Given that x-root 5 is a factor of cubic polynomial x^3- 3 root 5 x^2 + 13x - 3 root 5 , find all the zeroes of the polynomial

Posted by Mehal Jain 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

On factorising the quotient, we get
{tex}x ^ { 2 } - 2 \sqrt { 5 } x - 15 = x ^ { 2 } - 3 \sqrt { 5 } x + \sqrt { 5 }x-15{/tex}
{tex}= x ( x - 3 \sqrt { 5 } ) + \sqrt { 5 } ( x - 3 \sqrt { 5 } ){/tex}
{tex}= ( x + \sqrt { 5 } ) ( x - 3 \sqrt { 5 } ){/tex}
{tex}\therefore\ {/tex}{tex}( x + \sqrt { 5 } ) ( x - 3 \sqrt { 5 } ) = 0{/tex}
{tex}\Rightarrow\ x = - \sqrt { 5 } , 3 \sqrt { 5 }{/tex}
Therefore, all the zeroes are {tex}\sqrt { 5 } , - \sqrt { 5 } \text { and } 3 \sqrt { 5 }{/tex}.

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Find the zeros of the polynomial x*x+x-p(p+1).

Posted by Sunny Kumar 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

We have,
{tex}f(x) = x^2 + x - p(p + 1){/tex}
{tex}= x^2 + (p + 1)x - px - p(p + 1) {/tex}[As x =(p + 1)x - px ]
{tex}= x[ x + (p + 1)] - p[ x + (p + 1)]{/tex}
{tex}= ( x - p)[x + (p + 1)]{/tex}
{tex}{/tex} Now, f(x) = 0
{tex}\Rightarrow{/tex} ( x - p)[x + (p + 1)] = 0
{tex}\Rightarrow{/tex} x - p = 0 or x + (p +1) = 0
{tex}\Rightarrow{/tex} x = p or x = -(p + 1)
Thus, the required zeros of f(x) are p and -(p + 1).