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Ask QuestionPosted by Vineet Singh 4 years, 9 months ago
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Posted by Igl Praveen 4 years, 9 months ago
- 1 answers
Gaurav Seth 4 years, 9 months ago
Answer:
The value of p(-x)+p(x) is 6.
Step-by-step explanation:
The given function is
.... (1)
Substitute x=-x, it the given function.
..... (2)
Now add the equations (1) and (2).
Combined like terms.
Therefore the value of p(-x)+p(x) is 6.
Posted by Bhaduram Hembra 4 years, 9 months ago
- 4 answers
Posted by Devansh Jasuja 4 years, 9 months ago
- 1 answers
Posted by Swagata Lakshmi Mallik 4 years, 9 months ago
- 1 answers
Yogita Ingle 4 years, 9 months ago
Let AOB be the sector with O as center.
Given: Radius =r=5.2 cm
Perimeter of sector =16.4 cm
So, OA+AB+OB=16.4
⇒5.2+5.2+AB=16.4
⇒AB=6 cm
Area of sector =1/2 rl
=1/2×5.2×6
=15.6 cm2
Posted by Swagata Lakshmi Mallik 4 years, 9 months ago
- 2 answers
Yogita Ingle 4 years, 9 months ago
Radius = 10.5 cm and Sector angle = 60°
Perimeter of a sector of a circle = [(theta/360°)*2πr] + (2r)
⇒[(60/360)*2*22/7*10.5] + (2*10.5)
⇒ 1*462/6*7 + 21
⇒ 462/42 + 21
⇒ 11 + 21
⇒ 32 cm
So, the perimeter of sector is 32 cm
Posted by Najia Begum 4 years, 9 months ago
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Shardul Kadam 4 years, 9 months ago
Posted by Najia Begum 4 years, 9 months ago
- 3 answers
Igl Praveen 4 years, 9 months ago
Posted by Antara Saha 4 years, 9 months ago
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Нαяѕнιтα 🖤 4 years, 9 months ago
Posted by Shyni Manoj 4 years, 9 months ago
- 1 answers
Gaurav Seth 4 years, 9 months ago
Tan A = √2 - 1
Tan A = p / b { p: perpendicular , b:base}
h : hypotenuse
(√ 2 - 1 ) / 1 = p / b
p= √2 - 1 , b = 1
By pythagoras theorem ,
p² + b² = h²
(√2 - 1 )² + (1)² = h²
2 + 1 - 2√2 + 1 = h ²
h² = [4 - 2√2 ]
Sin A = p / h , Cos A = b / h
Sin A Cos A = ( p * b ) / h²
=> (√ 2 - 1 ) / ( 4 - 2 √2)
Rationalize Denomonator ,
=> ( √ 2 - 1 ) * ( 4 + 2 √2 ) / ( 4 - 2 √2 )(4 + 2 √2)
=> (4 √2 + 4 - 4 - 2√2 ) / ( 16 - 8 )
=> 2 √2 / 8
=> √2 / 4
Hence Proved
Posted by T. Srivatsa 4 years, 9 months ago
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Posted by Diksha Malik 4 years, 9 months ago
- 2 answers
Yogita Ingle 4 years, 9 months ago
In Δ ABC, B is at right angle.
Given, AB=24cm
BC=7cm
using Pythagoras theorem
AB2+BC2=AC2
⇒(24)2+(7)2=AC2
⇒AC=(242)+(7)2=576+49=625
⇒AC=25CM
sin A=BC/AC
= 7/25
Posted by Shresth Gupta 4 years, 9 months ago
- 3 answers
Posted by Yash Gaud 4 years, 9 months ago
- 1 answers
Siddhi Bhandari 4 years, 9 months ago
Posted by Sheetal Kumari 4 years, 9 months ago
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Posted by Anurag Kumar 4 years, 9 months ago
- 3 answers
Siddhi Bhandari 4 years, 9 months ago
Posted by Rajeevan Rajkumar 4 years, 9 months ago
- 1 answers
Gaurav Seth 4 years, 9 months ago
In the question, we have to find the radius (r) of the circle.
We are given that a piece of wire 20 cm long is bent into the form of an arc of a circle subtending an angle of 60 degrees at its centre. So, here the arc length will be the same as 20 cm, as shown in the figure below.
Next, we know the arc length formula as arclength=θ360o×(2πr)arclength=θ360o×(2πr), here θθ is the angle subtended at the centre of the circle. So we have: θ=60oθ=60o and arclength=20cmarclength=20cm. So using the formula we will find the radius of the circle, as shown below:
⇒arclength=θ360o×(2πr)⇒20=60o360o×(2πr)⇒20=16×(2πr)⇒60=(πr)⇒r=60π⇒arclength=θ360o×(2πr)⇒20=60o360o×(2πr)⇒20=16×(2πr)⇒60=(πr)⇒r=60π
So we get the required radius of the circle as r=60πcmr=60πcm.
Posted by Abhinav Gaur 4 years, 9 months ago
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Posted by K. Vishnu Vardhan Reddy 4 years, 9 months ago
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Posted by Aishwarya Marne 4 years, 9 months ago
- 2 answers
Gaurav Seth 4 years, 9 months ago
HCF of two positive integers can be find using the Euclid’s Division Lemma algorithm
We know that for any two integers a. b. we can write following expression
a=bq + r , 0≤r<b
If r=0 .then
HCF( a. b) =b
If r≠0 , then
HCF ( a. b) = HCF ( b.r)
Again expressing the integer b.r in Euclid’s Division Lemma, we get
b=pr + r1
HCF ( b,r)=HCF (r,r1)
Similarly successive Euclid’s division can be written until we get the remainder zero, the divisor at that point is called the HCF of the a and b
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HCF (a,b) =1 – Then a and b are co primes.
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Product of primes Theorem of Arithmetic – Composite Number = Product of Primes
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HCF and LCM by prime factorization method:
HCF = Product of the smallest power of each common factor in the numbers
LCM = Product of the greatest power of each prime factor involved in the number -
Important Formula HCF (a,b) X LCM (a,b) =a X b
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Important concept for rational Number – Terminating decimal expression can be written in the form p/2n5m
Gaurav Seth 4 years, 9 months ago
| S. No | Type of Numbers | Description |
| 1 | Natural Numbers | N = {1,2,3,4,5 > It is the counting numbers |
| 2 | Whole number | W= {0,1,2,3,4,5> It is the counting numbers + zero |
| 3 | Integers | All whole numbers including Negative number + Positive number ……-4,-3,-2,-1,0,1,2,3,4,5… so on. Like whole numbers, integers don’t include fractions or decimals. |
| 4 | Positive integers | Z+ = 1,2,3,4,5, …… |
| 5 | Negative integers | Z– = -1,-2,-3,-4,-5, …… |
| 6 | Rational Number | A number is called rational if it can be expressed in the form p/q where p and q are integers (q> 0). Ex: P/q, 4/5 |
| 7 | Irrational Number | A number is called rational if it cannot be expressed in the form p/q where p and q are integers (q> 0). Ex: √2, Pi, … etc |
| 8 | Real Numbers | A real number is a number that can be found on the number line. Real Numbers are the numbers that we normally use and apply in real-world applications. Real Numbers include Natural Numbers, Whole Numbers, Integers, Fractions, Rational Numbers and Irrational Numbers |
Posted by Raveena Jakhu 4 years, 9 months ago
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Posted by Aditi Vaishya 4 years, 9 months ago
- 1 answers
Posted by Francis Lalu 4 years, 9 months ago
- 2 answers
Posted by Francis Lalu 4 years, 9 months ago
- 1 answers
Tarannum Jahan 4 years, 9 months ago
Posted by Francis Lalu 4 years, 9 months ago
- 1 answers
Yogita Ingle 4 years, 9 months ago
Given A.P: 84,80,76,72,...
First term (a) = 84,
We have to find when the first negative term appear in the A.P.
Therefore,
First negative term appear at 23 rd term in A.P
Posted by Divyansh Trivedi 4 years, 9 months ago
- 1 answers
Posted by Shiva Karthik 4 years, 9 months ago
- 2 answers
Yogita Ingle 4 years, 9 months ago
1st term t1 = (k + 1).
2nd term t2 = 3k.
3rd term t3 = (4k + 2).
Given that t1,t2,t3 are in AP.
We know that when they are in AP, their common difference will be:
t2 - t1 = t3 - t2
3k - (k + 1) = (4k + 2) - 3k
3k - k - 1 = 4k + 2 - 3k
2k - 1 = k + 2
2k = k + 2 + 1
2k = k + 3
2k - k = 3
k = 3.
Therefore the value of k = 3.
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