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Ask QuestionPosted by Shambhu Kumar 4 years, 10 months ago
- 1 answers
Posted by Dhananjay Choudhary 4 years, 10 months ago
- 0 answers
Posted by Vicky Kumar 4 years, 10 months ago
- 3 answers
Supriya Mishra 4 years, 10 months ago
Posted by Akshay Kalia 4 years, 10 months ago
- 2 answers
Gaurav Seth 4 years, 10 months ago
The ratio is 3:4
A(2,5) B(x,y) and C divides the line at C(-1,2)
Step-by-step explanation:For X coordinate
-1=3*x+4*2/3+4
-1=3x+8/7
-7-8=3x
-15=3x
X=-5
For y coordinate :
2=3y+4*5/7
14=3y+20
14-20=3y
-6=3y
Y=-6/3
Y=-2
So x= -5 and y= -2
Posted by Sonu Jha 4 years, 10 months ago
- 2 answers
Posted by ಹರ್ಷನಂದ ಹರ್ಷನಂದ 4 years, 10 months ago
- 4 answers
ಹರ್ಷನಂದ ಹರ್ಷನಂದ 4 years, 9 months ago
ಹರ್ಷನಂದ ಹರ್ಷನಂದ 4 years, 10 months ago
Posted by Shreya Chatterjee 4 years, 10 months ago
- 1 answers
Posted by Tanmay Singh 4 years, 10 months ago
- 2 answers
Yogita Ingle 4 years, 10 months ago
38220>196 we always divide greater number with smaller one.
Divide 38220 by 196 then we get quotient 195 and no remainder so we can write it as
38220 = 196 * 195 + 0
As there is no remainder so deviser 196 is our HCF
Posted by Sathi Sardar 4 years, 10 months ago
- 2 answers
Gaurav Seth 4 years, 10 months ago
S o l u t i o n
Given sinA=3/4
Or p/h=3/4
Let p=3k and h=4k
Now By Pythagoras theorem
p2 + b2 =h2
9k2 + b2 =16k2
b2 = 7k2
Or b=+k√7
Now CosA=b/h=√7/4
Now tanA=SinA/CosA=3/√7
Posted by Niharika Raj 4 years, 10 months ago
- 3 answers
ಹರ್ಷನಂದ ಹರ್ಷನಂದ 4 years, 10 months ago
Devanshi Saxena 4 years, 10 months ago
Devanshi Saxena 4 years, 10 months ago
Posted by Naman Verma 4 years, 10 months ago
- 1 answers
Gaurav Seth 4 years, 10 months ago
O is the centre of the given circle.
A tangent PR has been drawn touching the circle at point P.
Draw QP ⊥ RP at point P, such that point Q lies on the circle.
∠OPR = 90° (radius ⊥ tangent)
Also, ∠QPR = 90° (Given)
∴ ∠OPR = ∠QPR
Now, above case is possible only when centre O lies on the line QP.
Hence, perpendicular at the point of contact to the tangent to a circle passes through the centre of the circle.
Posted by Anna Babu 4 years, 10 months ago
- 3 answers
Posted by Mđ Ąmãăņ 4 years, 10 months ago
- 1 answers
Gaurav Seth 4 years, 10 months ago
AB+CD=AD+BC
Proof :
AP=AS
PB=BQ
CQ=CR
RD=DS (the length of tangent drawn from an external point to a circle are equal)
AP + PB + RD + CR = AS + BQ + DS + CQ
=>AB + CD = AD + BC
Hence proved
Posted by Disha Jangid 4 years, 10 months ago
- 1 answers
Yogita Ingle 4 years, 10 months ago
Let the point on the x-axis be (x,0)
Distance between (x,0) and (7,6)=√(7−x)2+(6−0)2=√72+x2−14x+36 = √x2−14x+85
Distance between (x,0) and (−3,4)=√(−3−x)2+(4−0)2 = √32+x2+6x+16= √x2+6x+25
As the point (x,0) is equidistant from the two points, both the distances
calculated are equal.
x2−14x+85=x2+6x+25
=>x2−14x+85=x2+6x+25
85−25=6x+14x
60=20x
x=3
Thus, the point is (3,0)
Posted by Akshita Vibha Gupta 4 years, 10 months ago
- 4 answers
Posted by Nitin Singh 4 years, 10 months ago
- 1 answers
Gaurav Seth 4 years, 10 months ago
YES
After introducing two levels of mathematics papers for class 10 students, CBSE will now offer applied mathematics as an academic elective at the senior secondary level for those who do not want to take it up for higher studies or won't opt for engineering which require a broader understanding of the subject.
The elective subject aimed at developing an understanding of basic mathematical and statistical tools and their applications in the field of commerce and social science, will be offered as elective for class 11students from 2020 academic session and ultimately for class 12 students from the year after.
Students who had taken up basic mathematics in class 10 will be allowed to opt for applied mathematics at senior secondary level.
"Mathematics is widely used in higher studies — in the field of Economics, Commerce, Social Sciences and many others. It has been observed that the existing syllabus of mathematics aligns well with science subjects, but not so much with commerce or social science-based subjects in university education. Keeping this in mind, one more elective course in mathematics syllabus will be offered for senior secondary classes with an aim to provide students relevant experience in mathematics which can be used in the fields other than physical sciences," the Central Board of Secondary Education (CBSE) said in an official notification. According to CBSE officials, a course by this name was earlier designed as a skill subject.
Posted by Rani Kumari 4 years, 10 months ago
- 1 answers
Gaurav Seth 4 years, 10 months ago
The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
Solution:
Let the fixed charge be Rs x and per km charge be Rs y.
According to the question,
x + 10y = 105 …………….. (1)
x + 15y = 155 …………….. (2)
From (1), we get x = 105 – 10y ………………. (3)
Substituting the value of x in (2), we get
105 – 10y + 15y = 155
5y = 50
y = 10 …………….. (4)
Putting the value of y in (3), we get
x = 105 – 10 × 10 = 5
Hence, fixed charge is Rs 5 and per km charge = Rs 10
Charge for 25 km = x + 25y = 5 + 250 = Rs 255
Posted by Suhani Sendhav 4 years, 10 months ago
- 5 answers
Posted by Sagar Bhadu 4 years, 10 months ago
- 2 answers
Ayush Srivastav 4 years, 10 months ago
Posted by Kalyani Arun 4 years, 10 months ago
- 2 answers
Posted by Harshitha Harshitha 4 years, 10 months ago
- 1 answers
Posted by Divansh Jamwal 4 years, 10 months ago
- 2 answers
Ayush Srivastav 4 years, 10 months ago
Gaurav Seth 4 years, 10 months ago
let first term of Ap is a and common difference is d
use formula
tn=a+(n-1) d
now
t11 = a + (11-1)
d = a + 10d = 38
same way
t16 = a + 15d =73
solve this equation then
a= -32 and d= 7
now t31 = a+30d
= -32 + 210 = 178
Posted by Rohit Pawar 4 years, 10 months ago
- 2 answers
King Adithya H M... 4 years, 10 months ago
Yogita Ingle 4 years, 10 months ago
(Sin 30 + cos 30) - (sin 60 + Cos 60)
= [Sin30 + sin (90-30)] - [sin60 - sin (90 - 60 ) ]
= sin 30 + sin 60 - sin 60 - sin 30
=0
Posted by Mehak Kashyap 4 years, 10 months ago
- 2 answers
King Adithya H M... 4 years, 10 months ago
Yogita Ingle 4 years, 10 months ago
11×6+12×5+36
= 6(11× 1 + 2 × 5 + 6)
= 6(11 +10 + 6)
= 6 (27) = 162
Therefore according to fundamental theorem of arithmetic product of a composite and prime number is always composite number.
Posted by Ankit Sharma 4 years, 10 months ago
- 2 answers
Yogita Ingle 4 years, 10 months ago
to find the perimeter of the larger hemisphere- 2πr/2 = πr = 44 cm ( radius of larger hemisphere is 14 cm)
perimeters of two smaller hemispheres - 2×2πr /2 = 2πr = 44 cm ( radius of smaller hemispheres is 7 cm )
adding both we get 88cm as perimeter of the shaded region.
King Adithya H M... 4 years, 10 months ago
Posted by Sushant Sharma 4 years, 10 months ago
- 4 answers
King Adithya H M... 4 years, 10 months ago
Kush Mittal 4 years, 10 months ago
Posted by Sandeep Arya Gu 4 years, 10 months ago
- 2 answers
Supriya Mishra 4 years, 10 months ago
Gaurav Seth 4 years, 10 months ago
Let first number = a,
second number = b
a = 27 ( given )
HCF(a,b) = 9
and
LCM (a,b) = 459
To find:
Value of b
solution:
HCF×LCM= product of two numbers
⇒27 b = 9 × 459
⇒ b = 153
Posted by A B 4 years, 10 months ago
- 1 answers
Posted by Anjali Anjali 4 years, 10 months ago
- 2 answers
Yogita Ingle 4 years, 10 months ago
As per figure,
- Diagonal of the quadrant = radius =
or Side = 6 cm
- Area of Square =
= 2
=
- Area of Quadrant
of the Circle =
=
= 56.54
- Area of Shaded Region = Area of Quadrant - Area of Square
= 56.54 - 36
= 20.54
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