Ask questions which are clear, concise and easy to understand.
Ask QuestionPosted by Narvir Rana 5 years, 4 months ago
- 0 answers
Posted by Shruti Sharma 5 years, 4 months ago
- 1 answers
Yogita Ingle 5 years, 4 months ago
3x + y -11=0........... (i)
x - y -1=0
x = y + 1 ............ (ii)
Substitute (ii) in (i), we get
3(y+ 1) + y - 11 =0
3y + 3 + y - 11 =0
4y - 8 =0
4y = 8
y = 8/2 =4
Put y = 4 in (ii) , we get
x = y + 1
x= 5
Posted by Dark ..? 5 years, 4 months ago
- 4 answers
Posted by Nandita Chougule 5 years, 4 months ago
- 2 answers
? Pranali.A.P ? 5 years, 4 months ago
Posted by Mayank Tiwary 5 years, 4 months ago
- 1 answers
Posted by Gourinandana Sathyan 5 years, 4 months ago
- 2 answers
Arpita Singh 5 years, 4 months ago
Gaurav Seth 5 years, 4 months ago
HCF×LCM= product of two numbers
Let the other number be x
⇒27x=9×459
⇒x=153
Posted by Govind Mani 5 years, 4 months ago
- 1 answers
Lucifer?? Morningstar?? 5 years, 4 months ago
Posted by Sanjay Rawat 5 years, 4 months ago
- 2 answers
Lucifer?? Morningstar?? 5 years, 4 months ago
Lucifer?? Morningstar?? 5 years, 4 months ago
Posted by Sanam Singh 5 years, 4 months ago
- 2 answers
Oav Ka Rajkumar ??? 5 years, 4 months ago
Yogita Ingle 5 years, 4 months ago
Given series is 3, 8, 13, ...... 253
here, first term , a = 3
common difference , d = 5
Let us find total number of terms at first.
use
=> 253 = 3 + (n - 1)5
=> 250 = 5(n - 1)
=> n - 1 = 50
=> n = 51
so, there are 51 terms in given series.
now we know, mth term from last =last term - mth term + 1
so, 20th term from last = 51 - 20 + 1 = 32
hence, 20th from last = 32th term from first
use , T32= a+(32 -1)d
= 3 + 31 × 5 = 3 + 155 = 158
hence , 20th term from last = 158
Posted by Tridip Kharka 5 years, 4 months ago
- 0 answers
Posted by Shraddha Kunthe 5 years, 4 months ago
- 3 answers
Posted by Mr__Priyanshu ❤️______ __ 5 years, 4 months ago
- 1 answers
Yogita Ingle 5 years, 4 months ago
Let us assume that √5 is a rational number.
Sp it t can be expressed in the form p/q where p,q are co-prime integers and q≠0
⇒√5=p/q
On squaring both the sides we get,
⇒5=p²/q²
⇒5q²=p² —————–(i)
p²/5= q²
So 5 divides p
p is a multiple of 5
⇒p=5m
⇒p²=25m² ————-(ii)
From equations (i) and (ii), we get,
5q²=25m²
⇒q²=5m²
⇒q² is a multiple of 5
⇒q is a multiple of 5
Hence, p,q have a common factor 5. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number
√5 is an irrational number
Posted by Ashwini Kunthe 5 years, 4 months ago
- 1 answers
Gaurav Seth 5 years, 4 months ago
Let say AP is
a , a + d , a + 2d , a + 3d , a + 4d
Sn = (n/2)(2a + (n-1)d )
Sum of 5 terms
= (5/2)(2a + 4d) = 55
=> a + 2d = 11
fourth term is five more than the sum of the first two terms.
=> a + 3d = 5 + a + a + d
=> 2d - a = 5
Adding both
=> 4d = 16
=> d = 4
=> a = 3
AP
3 , 7 , 11 , 15 , 19
Posted by ?Suhani Arora? 5 years, 4 months ago
- 1 answers
Posted by Deepak Gaur 5 years, 4 months ago
- 1 answers
?Suhani Arora? 5 years, 4 months ago
Posted by Diya Rajput 5 years, 4 months ago
- 4 answers
Posted by Ankita Kumari 5 years, 4 months ago
- 2 answers
Posted by Gulshan Yadav 5 years, 4 months ago
- 0 answers
Posted by Hardik Kathuria 5 years, 4 months ago
- 1 answers
Posted by Adawn Jason Brown 5 years, 4 months ago
- 0 answers
Posted by Rohin Jaswal 5 years, 4 months ago
- 1 answers
Meghna Thapar 5 years, 4 months ago
A quadratic typically has the form of ax^2 + bx + c = 0 however the only requirement is that |a|>0, there is not factor of x to a power higher than 2 and no other function involved ie x^3 is not a quadratic. Notice also that x^2 + sin(x) +3 is not quadratic. In math, we define a quadratic equation as an equation of degree 2, meaning that the highest exponent of this function is 2. The standard form of a quadratic is y = ax^2 + bx + c, where a, b, and c are numbers and a cannot be 0. Examples of quadratic equations include all of these: y = x^2 + 3x + 1.
Posted by Gauri Nandana 5 years, 4 months ago
- 1 answers
Posted by Asha Mahesh 5 years, 4 months ago
- 1 answers
Posted by Rakesh Verma 5 years, 4 months ago
- 1 answers
Saksham Yadav 5 years, 4 months ago
Posted by Bhumika Bhatia 5 years, 4 months ago
- 3 answers
Gaurav Seth 5 years, 4 months ago
Second Method
Multiples of 4 lies between 10 and 250 are 12, 16, 20, ...., 248.
These numbers form an AP with a = 12 and d = 4.
Let number of three-digit numbers divisible by 4 be n, an = 248
⇒ a + (n - 1) d = 248
⇒ 12 + (n - 1) × 4 = 248
⇒4(n - 1) = 248
⇒ n - 1 = 59
⇒ n = 60
Gaurav Seth 5 years, 4 months ago
We know that First multiple of 4 that is greater than 10 is 12. Next will be 16.
Therefore, 12, 16, 20, 24, …
All these are divisible by 4 and thus, all these are terms of an A.P. with first term as 12 and common difference as 4.
When we divide 250 by 4, the remainder will be 2. Therefore, 250 − 2 = 248 is divisible by 4.
The series is as follows.
12, 16, 20, 24, …, 248
Let 248 be the nth term of this A.P.
a = 12
d = 4
an = 248
an = a + (n - 1) d
248 = 12 + (n - 1) × 4
236/4 = n - 1
59 = n - 1
n = 60
Therefore, there are 60 multiples of 4 between 10 and 250.
Posted by Nithin.S Yadawad 5 years, 4 months ago
- 0 answers
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
