Ask questions which are clear, concise and easy to understand.
Ask QuestionPosted by Sanjana Kumari 4 years, 8 months ago
- 0 answers
Posted by Manish Tiwari 4 years, 8 months ago
- 1 answers
Yogita Ingle 4 years, 8 months ago
Let O (x, y) is the point of the circle
if three given points A (3,-7) B (3,3) and C (6,-6)
we know distance between circumference and center is always same. i.e radius .
now,
{tex}OA^2=OB^2=OC^2{/tex}
{tex}OA^2=OB^2{/tex}
{tex}=>(x-3)^2+(y+7)^2=(x-3)^2+(y-3)^2{/tex}
{tex}=>(x-3)^2-(x-3)^2=(y-3)^2-(y+7)^2{/tex}
{tex}
=> 0=(2y+4)(3){/tex}
{tex}=> y= -2{/tex}
now again ,
{tex}OB^2=OC^2{/tex}
{tex}(x-3)^2+(y-3)^2=(x-6)^2+(y+6)^2{/tex}
put y=-2
{tex}=>(x-3)^2+(-2-3)^2=(x-6)^2+(-2+6)^2{/tex}
{tex}=>(x-3)^2-(x-6)^2=16-25{/tex}
{tex}=>(2x-9)(3)=-9{/tex}
{tex}=> 2x= -3+9=6{/tex}
=> x=3
hence center co-ordinate is (3,-2)
Posted by ?Khùšhì? * 4 years, 8 months ago
- 3 answers
Malaika Sharma 4 years, 8 months ago
Posted by Anushka ................. 4 years, 8 months ago
- 5 answers
Yogita Ingle 4 years, 8 months ago
When we are asked to calculate the distance between a point and any of the axes, we calculate the perpendicular distance between them.
Therefore, the point (-3,-4) is 3 units away from the y-axis and 4 units from the x-axis. This is because in case of either axes the co-ordinate of the other axis is zero.
Therefore the distance between the point P(-3,-4) and x-axis is 4 units.
Divanshu Rawat 4 years, 8 months ago
Ansh Sharma 4 years, 8 months ago
Posted by Baisa Hkm ?? 4 years, 8 months ago
- 5 answers
Tanu Choudhary # 4 years, 8 months ago
Tanu Choudhary # 4 years, 8 months ago
Yogita Ingle 4 years, 8 months ago
Given equations are
2x+3y=11−−−−(1)
2x−4y=−24−−−−(2)
Form (1)
2x+3y=11
⇒2x=11−3y
⇒x=211−3y−−−(3)
substituting x in(2)
2x−4y=−24
⇒2(211−3y)−4y=−24
⇒11−3y−4y=−24
⇒11−7y=−24
⇒7y=35
⇒y=35/7
⇒y=5.
putting y = 5 in (3)
x=[11−3(5)]/2
⇒x=[11−15]/2
⇒x=−4/2
∴x=−2.
Hence x = -2 and y = 5 is the solution of the
equation.
Now, we have to find m
y=mx+3 ∴m=−1
5=3(−2)+3
5−3=−2m⇒−2m=2
⇒m=−2/x=−1
Posted by Saba Yasmeen Kamdod 4 years, 8 months ago
- 1 answers
Posted by Kanak Yadav 4 years, 8 months ago
- 1 answers
Posted by Mehul Kedia 4 years, 8 months ago
- 1 answers
Posted by Ganesh Tote 4 years, 8 months ago
- 0 answers
Posted by Charvi Sangwan 4 years, 8 months ago
- 2 answers
Divanshu Rawat 4 years, 8 months ago
Yogita Ingle 4 years, 8 months ago
Let x be the number of correct answers and y be the number of wrong answers.
Since Jayanti answered 120 questions therefore, x+y=120 that is x=120−y..........(1)
Also, it is given that one mark is awarded for each correct answer while 1/2 mark is deducted for every wrong answer and Jayanti got 90 marks, therefore,
x−0.5y=90..........(2)
Substituting the value of equation 1 in equation 2:
120−y−0.5y=90
−1.5y=−30
y=20
Therefore,
x=120−20=100
Hence, she answered 100 questions correctly.
Posted by Kajal Kansal 4 years, 8 months ago
- 2 answers
King?Adithya H. M.. 4 years, 8 months ago
Gaurav Seth 4 years, 8 months ago
sum of the zeroes = (a-b) + (a+b) + a = 3a i.e. 3a = - b/a
or 3a=3
or a=1
product of the zeroes = (a-b)*(a+b)*a= a3-ab2 i.e. 1 - b2 = c/a (by putting the value a=1)
or b2 =2
or b = +-root2
Posted by Hilal . 4 years, 8 months ago
- 2 answers
Posted by Misty Chan 4 years, 8 months ago
- 5 answers
Gaurav Seth 4 years, 8 months ago
CBSE Class 10 Mathematics (041) - Deleted portion:
UNIT I-NUMBER SYSTEMS |
|
Chapter |
Topics |
REAL NUMBERS |
Euclid’s division lemma |
UNIT II-ALGEBRA |
|
Chapter |
Topics |
POLYNOMIALS |
Statement and simple problems on division algorithm for polynomials with real coefficients. |
PAIR OF LINEAR EQUATIONS IN TWO VARIABLES |
cross multiplication method |
QUADRATIC EQUATIONS |
Situational problems based on equations reducible to quadratic equations |
ARITHMETIC PROGRESSIONS |
Application in solving daily life problems based on sum to n terms |
UNIT III-COORDINATE GEOMETRY |
|
Chapter |
Topics |
COORDINATE GEOMETRY |
Area of a triangle |
UNIT IV-GEOMETRY |
|
Chapter |
Topics |
TRIANGLES |
Proof of the following theorems are deleted · The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. · In a triangle, if the square on one side is equal to sumof the squares on the other two sides, the angle opposite to the first side is a right angle. |
CIRCLES |
No Deletion |
CONSTRUCTIONS |
Construction of a triangle similar to a given triangle. |
UNIT V- TRIGONOMETRY |
|
Chapter |
Topics |
INTRODUCTION TO TRIGONOMETRY |
Motivate the ratios whichever are defined at 0o and 90o |
TRIGONOMETRIC IDENTITIES |
Trigonometric ratios of complementary angles |
HEIGHTS AND DISTANCES |
No deletion |
UNIT VI-MENSURATION |
|
AREAS RELATED TO CIRCLES |
Problems on central angle of 120° |
SURFACE AREAS AND VOLUMES |
Frustum of a cone |
UNIT VI-STATISTICS & PROBABILITY |
|
Chapter |
Topics |
STATISTICS |
· Step deviation Method for finding the mean · Cumulative Frequency graph |
PROBABILITY |
No deletion |
Posted by Sumit Rathi 4 years, 8 months ago
- 1 answers
Posted by Arati Punajiche 4 years, 8 months ago
- 2 answers
Mayank Singhal 4 years, 8 months ago
Mayank Singhal 4 years, 8 months ago
Posted by Logeshwari Lokeshwari.D 4 years, 8 months ago
- 2 answers
Yogita Ingle 4 years, 8 months ago
Letus assume that 3 + 2√5 is a rational number.
Soit can be written in the form a/b
3 + 2√5 = a/b
Here a and b are coprime numbers and b ≠ 0
Solving3 + 2√5 = a/b we get,
=>2√5 = a/b – 3
=>2√5 = (a-3b)/b
=>√5 = (a-3b)/2b
This shows (a-3b)/2b is a rational number. But we know that But √5 is an irrational number.
so it contradictsour assumption.
Our assumption of3 + 2√5 is a rational number is incorrect.
3 + 2√5 is an irrational number
Hence proved
Logeshwari Lokeshwari.D 4 years, 8 months ago
Posted by Pratyush Yadav 4 years, 8 months ago
- 0 answers
Posted by Tanu Choudhary # 4 years, 8 months ago
- 2 answers
Tanu Choudhary # 4 years, 8 months ago
Yogita Ingle 4 years, 8 months ago
A modem is a hardware device which means modulator-demodulator.
modems are used to transfer data from one computer network to another computer network through telephone lines.
the computer network works in digital mode, while analog technology is used for carrying messages across phone lines.
the modulator converts information from digital mode to analog mode at the transmitting end and demodulator converts the same from analog to digital at receiving end.
Posted by Krishna Verma 4 years, 8 months ago
- 4 answers
Ujjawal Mittal 4 years, 8 months ago
Posted by Krishna Verma 4 years, 8 months ago
- 1 answers
Tanu Choudhary # 4 years, 8 months ago
Posted by Vaishnavi. B Bolisetty 4 years, 8 months ago
- 2 answers
Mayank Singhal 4 years, 8 months ago
Tanu Choudhary # 4 years, 8 months ago
Posted by Baisa Hkm ?? 4 years, 8 months ago
- 3 answers
Mayank Singhal 4 years, 8 months ago
Posted by Fathima Fathima 4 years, 8 months ago
- 0 answers
Posted by Himanshu Yadav 4 years, 8 months ago
- 3 answers
Mayank Singhal 4 years, 8 months ago
Niral Dharawat 4 years, 8 months ago
Posted by Taran Tpk 4 years, 8 months ago
- 1 answers
Yogita Ingle 4 years, 8 months ago
Given that, the point P(2,1) lies on the line segment joining the points A(4,2) and 8(8, 4), which shows in the figure below:
Posted by Pihu ✨ 4 years, 8 months ago
- 1 answers
Yogita Ingle 4 years, 8 months ago
Here a=2,n=100
And d=7−2=5
Sn= n/2 [2a+(n−1)d]
S100 = 100/2 [2(2)+(100−1)5]
= 50[4 + 495 ]
= 50(499)
=24950
Posted by Charvi Sangwan 4 years, 8 months ago
- 2 answers
Yogita Ingle 4 years, 8 months ago
The conditions for each of the solutions is given by;
- If
, then there is a unique solution for the pair of linear equations.
- If
, then there are infinitely many solutions for the pair of linear equations.
- If
, then there is no solution for the pair of linear equations.
So, according to our question;
SO,
Here, the third condition is satisfying that; which means
.
Therefore, the given pair of linear equations has no solution.
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Sumesh ☺️☺️☺️ 4 years, 8 months ago
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