Ask questions which are clear, concise and easy to understand.
Ask QuestionPosted by Nisha Shree 2 years, 10 months ago
- 4 answers
Dilsad Alam 2 years, 10 months ago
Posted by Arvind Kumar 2 years, 10 months ago
- 3 answers
Preeti Dabral 2 years, 10 months ago
{tex}\text { Let } \alpha \text { and } \beta \text { are the zeroes of the polynomial } f(x)=a x^2+b x+c \text {. }{/tex}
{tex}\therefore \quad(\alpha+\beta)=\frac{-\mathbf{b}}{\mathbf{a}} and \alpha \beta=\frac{\mathbf{c}}{\mathbf{a}} According to the question, \frac{1}{\alpha} and \frac{1}{\beta} are the zeroes of the required quadratic polynomial{/tex}
Sum of zeroes of required polynomial
{tex} \begin{aligned} \mathbf{S}^{\prime} & =\frac{1}{\alpha}+\frac{1}{\beta} \\ & =\frac{\alpha+\beta}{\alpha \beta} \\ & =\frac{-\mathbf{b}}{\mathbf{c}} \end{aligned} [From equation (i) and (ii)] and product of zeroes of required polynomial =\frac{1}{\alpha} \times \frac{1}{\beta}. P^{\prime}=\frac{1}{\alpha \beta} =\frac{a}{c} [From equation (ii)]{/tex}
Equation of the required polynomial.
{tex} =\mathbf{k}\left(\mathbf{x}^2-\mathbf{S}^{\prime} \mathbf{x}+\mathbf{p}^{\prime}\right) where $k$ is any non-zero constant =k\left(x^2-\left(\frac{-b}{c}\right) x+\frac{a}{c}\right) [From equation (iii) and (iv)] =k\left(x^2+\frac{b}{c} x+\frac{a}{c}\right) {/tex}
Posted by Shanmuga Kani 2 years, 10 months ago
- 2 answers
Preeti Dabral 2 years, 10 months ago
The given equations are
{tex}71x + 37y = 253{/tex}.........(i)
{tex}37x + 71y = 287{/tex}.........(ii)
Adding (i) and (ii),we get
{tex}108x + 108y = 540{/tex}
{tex}108(x + y) = 540{/tex}
{tex}\therefore x + y = \frac { 540 } { 108 } = 5{/tex}......(iii)
Subtracting (ii) from (i),
{tex}34x - 34y = 253 - 287 = -34{/tex}
{tex}34(x - y) = -34{/tex}
{tex}\therefore x - y = - \frac { 34 } { 34 } = - 1{/tex}....(iv)
Adding (iii) and (iv)
2x = 5 - 1 = 4
{tex}\Rightarrow x = 2{/tex}
Subtracting (iv) from (iii)
2y = 5 + 1 = 6
{tex}\Rightarrow y = 3{/tex}
{tex}\therefore{/tex} Solution is x = 2, y = 3
Posted by Mehak Preet Kaur 2 years, 10 months ago
- 2 answers
Posted by Tripti Rawat 2 years, 10 months ago
- 5 answers
Posted by Sona Cutie 2 years, 10 months ago
- 0 answers
Posted by Sabari Karthikeyan 2 years, 10 months ago
- 5 answers
Posted by Nikita Kumari 2 years, 10 months ago
- 3 answers
Posted by Testing Testing1 2 years, 10 months ago
- 0 answers
Posted by Chandrima Choudhary 2 years, 10 months ago
- 0 answers
Posted by Shivam Yadav 2 years, 10 months ago
- 1 answers
Posted by Krish Kanyal 2 years, 10 months ago
- 2 answers
Aditya Kediya 2 years, 10 months ago
Posted by Ravi Teza 2 years, 10 months ago
- 2 answers
Posted by Palak Singh 2 years, 10 months ago
- 3 answers
Posted by Mahash Thennavan 2 years, 10 months ago
- 5 answers
Posted by Abhinav Dev 2 years, 10 months ago
- 3 answers
Mohammed Affiq 2 years, 10 months ago
Posted by Sameeksha Rawat 2 years, 10 months ago
- 5 answers
Posted by Kirti Rane 2 years, 10 months ago
- 1 answers
Preeti Dabral 2 years, 10 months ago
{tex}\begin{aligned} & \text { LHS }=\frac{\sin A-\sin B}{\cos A+\cos B}+\frac{\cos A-\cos B}{\sin A+\sin B} \\ & =\frac{\sin ^2 A-\sin ^2 B+\cos ^2 A-\cos ^2 B}{(\cos A+\cos B)(\sin A+\sin B)} \\ & =\frac{\left(\sin ^2 A+\cos ^2 A\right)-\left(\sin ^2 B+\cos ^2 B\right)}{(\cos A+\cos B)(\sin A+\sin B)} \\ & =(\cos A+\cos B)(\sin A+\sin B) \\ & =0=\text { RHS } \end{aligned}{/tex}
Posted by Bunny . 2 years, 10 months ago
- 1 answers
Preeti Dabral 2 years, 10 months ago
Let (-4, 6) divide AB internally in the ratio k:1. Using the section formula, we get
{tex}( - 4,6 ) = \left( \frac { 3 k - 6 } { k + 1 } , \frac { - 8 k + 10 } { k + 1 } \right){/tex}
So, {tex}- 4 = \frac { 3 k - 6 } { k + 1 }{/tex}
i.e., -4k - 4 = 3k - 6
i.e., 7k = 2
i.e., k:1 = 2:7
The same can be checked for the y-coordinate also.
Therefore, the ratio in which the point (-4,6) divides the line segment AB is 2: 7.
Posted by Aniket Topwal 2 years, 10 months ago
- 1 answers
Preeti Dabral 2 years, 10 months ago
Let A(a, a), B(-a, -a) C{tex}\left( { - \sqrt 3 a,,\sqrt 3 a} \right){/tex}
{tex}AB = \sqrt {{{( - a - a)}^2} + {{( - a - a)}^2}} = \sqrt {8{a^2}} = 2\sqrt 2 a{/tex}
{tex}BC = \sqrt {{{( - \sqrt 3 a + a)}^2} + {{(\sqrt 3 a + a)}^2}} = 2\sqrt 2 a{/tex}
{tex}AC = \sqrt {{{( - \sqrt 3 a - a)}^2} + {{(\sqrt 3 a - a)}^2}} = 2\sqrt 2 a{/tex}
{tex}\therefore {/tex} AB = BC = AC = {tex}2\sqrt 2 a{/tex}
{tex}ar\Delta ABC = \frac{{\sqrt 3 }}{4} \times {(side)^2}{/tex}
{tex} = \frac{{\sqrt 3 }}{4} \times \left( {2\sqrt 2 a} \right)^2{/tex}
{tex} = 2\sqrt 3 {a^2}{/tex} sq. units
Posted by Ritu Kumari 2 years, 10 months ago
- 2 answers
Posted by Vis H 2 years, 10 months ago
- 1 answers
Preeti Dabral 2 years, 10 months ago
Given: A circle with centre O. A tangent CD at C.
Diameter AB is produced to D.
BC and AC chords are joined, ∠BAC = 30°
To prove: BC = BD
Proof: DC is tangent at C and, CB is chord at C.
Therefore, ∠DCB = ∠BAC [∠s in alternate segment of a circle]
⇒ ∠DCB = 30° …(i) [∵ ∠BAC = 30° (Given)]
AOB is diameter. [Given]
Therefore, ∠BCA = 90° [Angle in s semi circle]
Therefore, ∠ABC = 180° - 90° - 30° = 60°
In ΔBDC,
Exterior ∠B = ∠D + ∠BCD
⇒ 60° = ∠D + 30°
⇒ ∠D = 30° …(ii)
Therefore, ∠DCB = ∠D = 30° [From (i), (ii)]
⇒ BD = BC [∵ Sides opposite to equal angles are equal in a triangle]
Hence, proved.
Posted by Aishiki Jana 2 years, 10 months ago
- 1 answers
Pranjal Gupta 2 years, 10 months ago
myCBSEguide
Trusted by 1 Crore+ Students
Test Generator
Create papers online. It's FREE.
CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
Rdm 😈 2 years, 10 months ago
0Thank You