Ask & Answer

Ask & Answer: India's Largest Education Community

1000+ExpertsQuick ResponsesReliable Answers

Your Question

Edit

Add relevant tags to get quick responses. Cancel Post

All Questions

New Question

9 months ago

Exam MUJEE

Has the MUJEE 2025 exam already conducted?

0 Follower 2 Views

Saurabh Khanduri

Contributor-Level 10

No, the MUJEE 2025 exam is still ongoing the exact date of the MUJEE 2025 exam has not been announced yet. Candidates who are preparing for the MUJEE 2025 are advised to regularly check the official website of Mangalayatan University to remain updated on the timeline for the MUJEE 2025 exam.

0 0

New Question

9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Ten Exemplar Solution

If the intercept of a line between the coordinate axes is divided by the point (–5, 4) in the ratio 1: 2, then find the equation of the line.

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t a a n d b b e t h e o n t h e g i v e n l i n e . \\ ∴ C o o r d i n a t e s o f A a n d B a r e (a, 0) a n d (0, b) r e s p e c t i v e l y \\ ∴ - 5 = \frac{1 \times 0 + 2 \times a}{1 + 2} \\ \Rightarrow 2 a = - 15 \Rightarrow a = \frac{- 15}{2} [\begin{array}{l} ? X = \frac{m_{1} x_{2} + m_{2} x_{1}}{m_{1} + m_{2}} \\ a n d Y = \frac{m_{1} y_{2} + m_{2} y_{1}}{m_{1} + m_{2}} \end{array}] \\ ∴ A = (\frac{- 15}{2}, 0) \\ a n d 4 = \frac{1 \times b + 0 \times 2}{1 + 2} \\ \Rightarrow 4 = \frac{b}{3} \Rightarrow b = 12 \\ ∴ B = (0, 12) \\ S o, t h e e q u a t i o n o f l i n e A B i s \\ y - y_{1} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} (x - x_{1}) \\ \Rightarrow y - 0 = \frac{12 - 0}{0 + \frac{15}{2}} (x + \frac{15}{2}) \\ \Rightarrow y = \frac{12 \times 2}{15} (x + \frac{15}{2}) \\ \Rightarrow y = \frac{8}{5} (x + \frac{15}{2}) \\ \Rightarrow 5 y = 8 x + 60 \\ \Rightarrow 8 x - 5 y + 60 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s 8 x - 5 y + 60 = 0 . \end{array}$

0 0

New Question

9 months ago

Illinois Institute of Technology Study Abroad

What are the overall subject rankings at Illinois Institute of Technology?

0 Follower 4 Views

Tasneem Hoda

Contributor-Level 10

Illinois Institute of Technology is ranked well for some of the courses such as Aerospace Engineering, Chemical Engineering, Architecture, Business and Management Studies, Law etc. The subject rankings at Illinois Tech are given in the table below:

Course Name	Shiksha Popularity Rank	US News Rank
MS	#94	#37
Aerospace Engineering	#37	#37
Chemical Engineering	#68	#67
Architecture and Built Environment	#29	NA
Biomedical Engineering	#95	#90

0 0

New Question

9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Ten Exemplar Solution

For what values of a and b the intercepts cut off on the coordinate axes by the line ax + by + 8 = 0 are equal in length but opposite in signs to those cut off by the line 2x – 3y + 6 = 0 on the axes.

0 Follower 5 Views

Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} T h e g i v e n e q u a t i o n a r e a x + b y + 8 = 0 \dots (i) \\ a n d 2 x - 3 y + 6 = 0 \dots (i i) \\ F r o m e q n . (i) w e g e t, \\ a x + b y + 8 = 0 \Rightarrow a x + b y = - 8 \\ \Rightarrow \frac{a}{- 8} x + \frac{b}{- 8} y = 1 \Rightarrow \frac{x}{\frac{- 8}{a}} + \frac{y}{\frac{- 8}{b}} = 1 \\ S o, t h e o n t h e a x e s a r e \frac{- 8}{a} a n d \frac{- 8}{b} \\ F r o m e q n . (i i) w e g e t, \\ 2 x - 3 y + 6 = 0 \Rightarrow 2 x - 3 y = - 6 \\ \Rightarrow \frac{2 x}{- 6} - \frac{3 y}{- 6} = 1 \Rightarrow \frac{x}{- 3} + \frac{y}{2} = 1 \\ S o, t h e o n t h e a x e s a r e - 3 a n d 2 . \\ A c c o r d i n g t o t h e q u e s t i o n \\ \frac{- 8}{a} = + 3 \Rightarrow a = - \frac{8}{3} \\ a n d \frac{- 8}{b} = - 2 \Rightarrow b = + 4 \\ H e n c e, t h e r e q u i r e d v a l u e s o f a a n d b a r e - \frac{8}{3} a n d 4 . \end{array}$

0 0

New Question

9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Ten Exemplar Solution

Find the equation of the line passing through the point of intersection of 2x + y = 5 and x + 3y + 8 = 0 and parallel to the line 3x + 4y = 7.

0 Follower 5 Views

Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$(a) G i v e n$

$\begin{array}{l} t h a t : 2 x + y = 5 \dots (i) \\ x + 3 y + 8 = 0 \dots (i i) \\ 3 x + 4 y = 7 \dots (i i i) \\ E q u a t i o n o f a n y l i n e t h r o u g h t h e o f o f e q n . (i) a n d e q n . (i i) i s \\ (2 x + y - 5) + λ (x + 3 y + 8) = 0 \dots (i v) [λ =] \\ \Rightarrow 2 x + y - 5 + λ x + 3 λ y + 8 λ = 0 \\ \Rightarrow (2 + λ) x + (1 + 3 λ) y - 5 + 8 λ = 0 \\ S l o p e o f l i n e m_{1} (s a y) = - \frac{(2 + λ)}{1 + 3 λ} [? m = \frac{- a}{b}] \\ S l o p e o f l i n e 3 x + 4 y = 7 i s \\ m_{2} (s a y) = - \frac{3}{4} \\ I f e q n . (i i i) i s p a r a l l e l t o e q n . (i v) t h e n \\ m_{1} = m_{2} \\ ∴ - \frac{(2 + λ)}{1 + 3 λ} = - \frac{3}{4} \\ \Rightarrow \frac{2 + λ}{1 + 3 λ} = \frac{3}{4} \Rightarrow 8 + 4 λ = 3 + 9 λ \\ \Rightarrow 9 λ - 4 λ = 5 \Rightarrow 5 λ = 5 \Rightarrow λ = 1 \\ O n p u t t i n g t h e v a l u e o f λ i n e q n . (i v) w e g e t \\ (2 x + y - 5) + 1 (x + 3 y + 8) = 0 \\ \Rightarrow 2 x + y - 5 + x + 3 y + 8 = 0 \\ \Rightarrow 3 x + 4 y + 3 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s 3 x + 4 y + 3 = 0 . \end{array}$

0 0

New Question

9 months ago

Mangalayatan University Colleges in Aligarh

How many job opportunities are made available to MBA students at Mangalayatan University?

0 Follower 1 View

Saurabh Khanduri

Contributor-Level 10

The exact figure for job opportunities during placements of MBA students is not available. However, in the year 2023, students of Mangalayatan University received 1000 job opportunities from 500 recruiters. The average median package for MBA students of Mangalayatan University stood at around INR 4.8 LPA while the highest package was around INR 10 LPA.

0 0

New Question

9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Ten Exemplar Solution

Find the equation of lines passing through (1, 2) and making an angle 30° with the y-axis.

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t t h e l i n e m a k e s a n g l e 3 0^{0} w i t h y a x i s \\ ∴ A n g l e m a d e b y t h e l i n e w i t h x a x i s i s 6 0^{0} \\ ∴ S l o p e o f t h e l i n e \\ m = t a n 6 0^{0} \\ \Rightarrow m = \sqrt[]{3} \\ S o, t h e e q u a t i o n o f t h e l i n e t h r o u g h t h e (1, 2) a n d s l o p e \sqrt[]{3} i s \\ y - y_{1} = m (x - x_{1}) \\ \Rightarrow y - 2 = \sqrt[]{3} (x - 1) \\ \Rightarrow y - 2 = \sqrt[]{3} x - \sqrt[]{3} \\ \Rightarrow y - \sqrt[]{3} x - \sqrt[]{3} - 2 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n o f l i n e i s y - \sqrt[]{3} x - \sqrt[]{3} - 2 = 0 . \end{array}$

0 0

New Question

9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Ten Exemplar Solution

Show that the tangent of an angle between the lines x/a + y/b = 1 and x/a − y/b = 1 is 2ab / (a² − b²).

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : \frac{x}{a} + \frac{y}{b} = 1 \dots (i) \\ a n d \frac{x}{a} - \frac{y}{b} = 1 \dots (i i) \\ S l o p e o f t h e e q n . (i) m_{1} (s a y) = - \frac{b}{a} \\ a n d s l o p e o f t h e e q n . (i i) m_{2} (s a y) = \frac{b}{a} \\ L e t θ b e t h e a n g l e b e t q w e e n t h e t w o g i v e n l i n e s \\ ∴ t a n θ = | \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} | = | \frac{- \frac{b}{a} - \frac{b}{a}}{1 + (- \frac{b}{a}) (\frac{b}{a})} | = | \frac{\frac{- 2 b}{a}}{1 - \frac{b^{2}}{a^{2}}} | = | \frac{- 2 a b}{a^{2} - b^{2}} | \\ \Rightarrow t a n θ = \frac{2 a b}{a^{2} - b^{2}} . H e n c e p r o v e d . \end{array}$

0 0

New Question

9 months ago

IIT Bhubaneswar - Indian Institute of Technology

What is the cut off of IIT Bhubaneswar?

0 Follower 8 Views

Saakshi Garg

Contributor-Level 6

IIT Bhubaneswar JEE Advanced cutoff 2025 was declared for admission to BTech course and its various specialisations. Students belonging to the All India quota can avail admission at IIT Bhubaneswar based on the JEE Advanced counselling results.

For the General AI category, the overall cutoff ranks were between 3785 and 16268. Hence, students need to secure at least a rank of 16268 or ideally lower to be eligible for admissions. Students can refer to the table below to know the round 1 and last round cutoff ranks for this category.

Course	Round 1 (Closing Rank)	Last Round (Closing Rank)
B.Tech. in Civil Engineering	13957	16156
B.Tech. in Electrical Engineering	7242	8165
B.Tech. in Mechanical Engineering	10078	11678
B.Tech. in Computer Science and Engineering	3785	4162
B.Tech. in Metallurgical and Materials Engineering	13695	16268
B.Tech. in Electronics and Communication Engineering	5535	5885
B.Tech. in Engineering Physics	10494	12466
B.Tech. in Mathematics and Computing	4607	4708

Please note that this is the cutoff information for the General AI category students, other

...more

0 0

New Question

9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Ten Exemplar Solution

Find the points on the line x + y = 4 which lie at a unit distance from the line 4x + 3y = 10.

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t (x_{1}, y_{1}) b e a n y l y i n g i n t h e e q u a t i o n x + y = 4 \\ ∴ x_{1} + y_{1} = 4 \dots (i) \\ o f t h e (x_{1}, y_{1}) f r o m t h e e q u a t i o n 4 x + 3 y = 10 \\ \frac{4 x_{1} + 3 y_{1} - 10}{\sqrt[]{{(4)}^{2} + {(3)}^{2}}} = 1 \\ | \frac{4 x_{1} + 3 y_{1} - 10}{5} | = 1 \\ 4 x_{1} + 3 y_{1} - 10 = \pm 5 \\ T a k i n g (+) s i g n 4 x_{1} + 3 y_{1} - 10 = 5 \\ \Rightarrow 4 x_{1} + 3 y_{1} = 15 \dots (i i) \\ F r o m e q n . (i) w e g e t y_{1} = 4 - x_{1} \\ P u t t i n g t h e v a l u e o f y_{1} i n e q n . (i i) w e g e t \\ 4 x_{1} + 3 (4 - x_{1}) = 15 \\ \Rightarrow 4 x_{1} + 12 - 3 x_{1} = 15 \\ \Rightarrow x_{1} + 12 = 15 \\ \Rightarrow x_{1} = 3 a n d y_{1} = 4 - 3 = 1 \\ S o, t h e r e q u i r e d i s (3, 1) \\ N o w t a k i n g (-) s i g n, 4 x_{1} + 3 y_{1} - 10 = - 5 \\ \Rightarrow 4 x_{1} + 3 y_{1} = 5 \dots (i i i) \\ F r o m e q n . (i) w e g e t y_{1} = 4 - x_{1} \\ 4 x_{1} + 3 (4 - x_{1}) = 5 \\ \Rightarrow 4 x_{1} + 12 - 3 x_{1} = 5 \\ \Rightarrow x_{1} + 12 = 5 \\ \Rightarrow x_{1} = - 7 a n d y_{1} = 4 - (- 7) = 11 \\ S o, t h e r e q u i r e d i s (- 7, 11) \\ H e n c e, t h e r e q u i r e d o n t h e g i v e n l i n e a r e (3, 1) a n d (- 7, 11) . \end{array}$

0 0

New Question

9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Ten Exemplar Solution

Find the equation of the lines which passes through the point (3, 4) and cuts off intercepts from the coordinate axes such that their sum is 14.

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} E q u a t i o n o f l i n e h a v i n g a a n d b o n t h e a x i s i s \\ \frac{x}{a} + \frac{y}{b} = 1 \dots (i) \\ G i v e n t h a t a + b = 14 \Rightarrow b = 14 - a \\ \Rightarrow \frac{x}{a} + \frac{y}{14 - a} = 1 \dots (i i) \\ I f e q n . (i i) p a s s e s t h r o u g h t h e (3, 4) t h e n \\ \frac{3}{a} + \frac{4}{14 - a} = 1 \\ \Rightarrow \frac{3 (14 - a) + 4 a}{a (14 - a)} = 1 \\ \Rightarrow 42 + a = 14 a - a^{2} \\ \Rightarrow a^{2} + a - 14 a + 42 = 0 \\ \Rightarrow a^{2} - 13 a + 42 = 0 \\ \Rightarrow a^{2} - 7 a - 6 a + 42 = 0 \\ \Rightarrow a (a - 7) - 6 (a - 7) = 0 \\ \Rightarrow (a - 6) (a - 7) = 0 \Rightarrow a = 6, 7 \\ ∴ b = 14 - 6 = 8, b = 14 - 7 = 7 \\ H e n c e, t h e r e q u i r e d e q u a t i o n o f l i n e s a r e \\ \frac{x}{6} + \frac{y}{8} = 1 \Rightarrow 4 x + 3 y = 24 \\ a n d \frac{x}{7} + \frac{y}{7} = 1 \Rightarrow x + y = 7 \end{array}$

0 0

New Question

9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Eleven Exemplar Solution

Find the equation of the circle which touches the $x$ -axis and whose center is $(1, 2)$ .

0 Follower 3 Views

Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} S i n c e t h e c i r c l e w h o s e c e n t r e i s (1, 2) t o u c h x - a x i s \\ ∴ r = 2 \\ S o, t h e e q u a t i o n o f t h e c i r c l e i s \\ {(x - h)}^{2} + {(y - k)}^{2} = r^{2} \\ \Rightarrow {(x - 1)}^{2} + {(y - 2)}^{2} = {(2)}^{2} \\ \Rightarrow x^{2} - 2 x + 1 + y^{2} - 4 y + 4 = 4 \\ \Rightarrow x^{2} + y^{2} - 2 x - 4 y + 1 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s x^{2} + y^{2} - 2 x - 4 y + 1 = 0 . \end{array}$

0 0

New Question

9 months ago

Law LLB

What is the future scope of LLB?

0 Follower 3 Views

Akansha Bisht

Contributor-Level 10

After pursuing an LLB, one can either pursue higher studies by doing an LLM or a PhD. This will open doors to academia and research. One may start practicing as an advocate by enrolling in the local state bar and also appearing in the All India Bar Examination. Candidates can also appear for various judicial and civil service examinations or other government and banking examinations.

0 0

New Question

9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Ten Exemplar Solution

Find the angle between the lines y = (2 – 3)(x + 5) and y = (2 + 3)(x – 7).

0 Follower 5 Views

Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} T h e g i v e n e q u a t i o n s a r e y = (2 - \sqrt[]{3}) (x + 5) \dots (i) \\ a n d y = (2 + \sqrt[]{3}) (x - 7) \dots (i i) \\ S l o p e o f t h e e q n . (i) m_{1} (s a y) = (2 - \sqrt[]{3}) \\ a n d s l o p e o f t h e e q n . (i i) m_{2} (s a y) = (2 + \sqrt[]{3}) \\ L e t θ b e t h e a n g l e b e t q w e e n t h e t w o g i v e n l i n e s \\ ∴ t a n θ = | \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} | = | \frac{2 - \sqrt[]{3} - 2 - \sqrt[]{3}}{1 + (2 - \sqrt[]{3}) (2 + \sqrt[]{3})} | = | \frac{- 2 \sqrt[]{3}}{2} | = | - \sqrt[]{3} | \\ \Rightarrow t a n θ = \sqrt[]{3} o r - \sqrt[]{3} \\ ∴ θ = 6 0^{0} o r 12 0^{0} \\ H e n c e, t h e r e q u i r e d a n g l e i s 6 0^{0} o r 12 0^{0} . \end{array}$

0 0

New Question

9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Eleven Exemplar Solution

If a circle passes through the points $(0, 0), (a, 0), (0, b)$ , then find the coordinates of its center.

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n points a r e (0, 0), (a, 0) a n d (0, b) \\ G e n e r a l e q u a t i o n o f t h e c i r c l e i s \\ x^{2} + y^{2} + 2 g x + 2 f y + c = 0 w h e r e t h e c e n t r e i s (- g, - f) a n d r a d i u s = \sqrt[]{g^{2} + f^{2} - c} \\ I f i t p a s s e s t h r o u g h (0, 0) \\ ∴ c = 0 \\ I f i t p a s s e s t h r o u g h (a, 0) a n d (0, b) t h e n \\ a^{2} + 2 g a + c = 0 \Rightarrow a^{2} + 2 g a = 0 [? c = 0] \\ ∴ g = - \frac{a}{2} \\ a n d 0 + b^{2} + 0 + 2 f b + c = 0 \Rightarrow b^{2} + 2 f b = 0 [? c = 0] \\ \Rightarrow f = - \frac{b}{2} \\ H e n c e, t h e c o o r d i n a t e s o f c e n t r e o f t h e c i r c l e a r e (- g, - f) = (\frac{a}{2}, \frac{b}{2}) \end{array}$

0 0

New Question

9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Ten Exemplar Solution

Find the equation of the line passing through the point (5, 2) and perpendicular to the line joining the points (2, 3) and (3, –1).

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$S l o p e o f t h e l i n e j o i n i n g t h r o u g h t h e (2, 3) a n d (3, - 1) i s \frac{- 1 - 3}{3 - 2} = - 4 S l o p e o f t h e r e q u i r e d l i n e w h i c h i s p e r p e n d i c u l a r t o i t = \frac{- 1}{- 4} = \frac{1}{4} [$
$m_{1} m_{2} = - 1] E q u a t i o n o f t h e l i n e t h r o u g h t h e (5, 2) i s y - 2 = \frac{1}{4} (x - 5) [$
$\begin{array}{l} y - y_{1} = m (x - x_{1})] \\ \Rightarrow 4 y - 8 = x - 5 \\ \Rightarrow x - 4 y + 3 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s x - 4 y + 3 = 0 . \end{array}$

0 0

New Question

9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Eleven Exemplar Solution

Show that the point $(x, y)$ given by $x = \frac{2 a t}{1 + t^{2}}$ and $y = \frac{a (1 - t^{2})}{1 + t^{2}}$ lies on a circle for all real values of $t$ , such that $- 1 < t < 1$ , where $a$ is any given real number.

0 Follower 5 Views

Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : x = \frac{2 a t}{1 + t^{2}} a n d y = \frac{a (1 - t^{2})}{1 + t^{2}} \\ \Rightarrow x^{2} + y^{2} = {(\frac{2 a t}{1 + t^{2}})}^{2} + {(\frac{a (1 - t^{2})}{1 + t^{2}})}^{2} \\ = \frac{4 a^{2} t^{2}}{{(1 + t^{2})}^{2}} + \frac{a^{2} {(1 - t^{2})}^{2}}{{(1 + t^{2})}^{2}} = \frac{4 a^{2} t^{2} + a^{2} (1 + t^{4} - 2 t^{2})}{{(1 + t^{2})}^{2}} \\ = \frac{4 a^{2} t^{2} + a^{2} + a^{2} t^{4} - 2 a^{2} t^{2}}{{(1 + t^{2})}^{2}} = \frac{a^{2} + a^{2} t^{4} + 2 a^{2} t^{2}}{{(1 + t^{2})}^{2}} \\ = \frac{a^{2} (1 + t^{4} + 2 t^{2})}{{(1 + t^{2})}^{2}} = \frac{a^{2} {(1 + t^{2})}^{2}}{{(1 + t^{2})}^{2}} = a^{2} \\ ∴ x^{2} + y^{2} = a^{2} w h i c h i s t h e e q u a t i o n o f a c i r c l e . \\ H e n c e, t h e g i v e n points l i e o n a c i r c l e . \end{array}$

0 0

New Question

9 months ago

Mangalayatan University Colleges in Aligarh

When does the admission process begin for an MBA at Mangalayatan University?

0 Follower 1 View

Saurabh Khanduri

Contributor-Level 10

The Mangalayatan University MBA course admission process begins in January every year. Candidates can enrol for the CUET-PG exam during this month. The exam is conducted in the month of March and the results are declared in the month of May.

After the results are announced, candidates can visit the official website of Mangalayatan University to apply for admission.

0 0

New Question

9 months ago

Merit Swiss Asian School of Hotel Management Panache Academy, Goa

Are admissions to UG Diploma at Panache Academy, Goa merit-based?

0 Follower 2 Views

Shikha Arora

Contributor-Level 10

Admission to UG Diploma course is based on merit at Panache Academy, Goa. The eligibility criteria for this course range between 17 years to 27 years. Panache Academy Goa popular programmes includes various BBA and Diplomas in several specialisations such as Airlines, Tourism, Hospitality Management for the duration of 1 year. Similarly, the Institute offers BBA in various specialisations for the duration of 3 years. For courses such as BBA, Panache Academy Goa considers an annual intake of 160 students at a total tuition fees of INR 1.9 Lakh.

0 0

New Question

9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Ten Exemplar Solution

Find the equation of the straight line which passes through the point (1, –2) and cuts off equal intercepts from axes.

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} I n t e r c e p t f o r m o f s t r a i g h t l i n e \\ \frac{x}{a} + \frac{y}{b} = 1, w h e r e a a n d b a r e t h e o n t h e a x i s \\ G i v e n t h a t a = b \\ ∴ \frac{x}{a} + \frac{y}{a} = 1 \dots (i) \\ I f e q n . (i) p a s s e s t h r o u g h t h e (1, - 2), w e g e t \\ \frac{1}{a} - \frac{2}{a} = 1 \Rightarrow - \frac{1}{a} = 1 \Rightarrow a = - 1 \\ S o, t h e e q u a t i o n o f t h e s t r a i g h t l i n e i s \\ \frac{x}{- 1} + \frac{y}{- 1} = 1 \Rightarrow x + y = - 1 \Rightarrow x + y + 1 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s x + y + 1 = 0 . \end{array}$

0 0

Ask & Answer
Panel of Experts

View Experts Panel

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Ask & Answer: India's Largest Education Community

Need guidance on career and education? Ask our experts

Your Question

All Questions

Share Your College Life Experience