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Ask QuestionPosted by Harman Preet 5 years, 2 months ago
- 3 answers
Sarthak Gupta 5 years, 2 months ago
Posted by Soniya Rana 5 years, 2 months ago
- 1 answers
Yogita Ingle 5 years, 2 months ago
Put n =1
put n =2
put n =3
So, A.P. become s: -1 , -4 , -7, ........
So, first term =a= -1
Common difference d = -4-(-1)=-7-(-4)= -3
Formula of sum of first n terms :
Put n =25
Hence the sum of first 25 terms of an AP is -925
Posted by Harshit Rajput 5 years, 2 months ago
- 1 answers
Yogita Ingle 5 years, 2 months ago
9y² - 12y + 2 = 0
(3y)² - 2(3y)(2) + 2² - 2² + 2 = 0
(3y - 2)² - 4 + 2 = 0
(3y - 2)² = 2
(3y - 2) = √2
taking (+ve)
3y = √2 + 2
y = (√2 + 2)/3
taking (-ve)
3y = -√2 + 2
y = (2 - √2)/2
Posted by Manoj Rakesh 5 years, 2 months ago
- 5 answers
Posted by Palak Jain 5 years, 2 months ago
- 2 answers
Nidhi Jain 5 years, 2 months ago
Posted by Meghana B L 5 years, 2 months ago
- 1 answers
Gaurav Seth 5 years, 2 months ago
For a quadratic equation ax² + bx + c =0, the term b² - 4ac is called discriminant (D) of the quadratic equation because it determines whether the quadratic equation has real roots or not ( nature of roots).
D= b² - 4ac
So a quadratic equation ax² + bx + c =0, has
i) Two distinct real roots, if b² - 4ac >0 , then x= -b/2a + √D/2a &x= -b/2a - √D/2a
ii) Two equal real roots, if b² - 4ac = 0 , then x= -b/2a or -b/2a
iii) No real roots, if b² - 4ac <0
-----------------------------------------------------------------------------------------------------
Solution:
i)
x² – 3x + 5 = 0
Comparing it with ax² + bx + c = 0, we get
a = 2, b = -3 and c = 5
Discriminant (D) = b² – 4ac
⇒ ( – 3)2 – 4 (2) (5) = 9 – 40
⇒ – 31<0
As b2 – 4ac < 0,
Hence, no real root is possible .
(ii) 3x² – 4√3x + 4 = 0
Comparing it with ax² + bx + c = 0, we get
a = 3, b = -4√3 and c = 4
Discriminant(D) = b² – 4ac
⇒ (-4√3)2 – 4(3)(4)
⇒ 48 – 48 = 0
As b² – 4ac = 0,
Hence, real roots exist & they are equal to each other.
the roots will be –b/2a and –b/2a.
-b/2a = -(-4√3)/2×3 = 4√3/6 = 2√3/3
multiplying the numerator & denominator by √3
(2√3) (√3) / (3)(√3) = 2 ×3 / 3 ×√3 = 2/√3
Hence , the equal roots are 2/√3 and 2/√3.
Posted by Tom Cruz 5 years, 2 months ago
- 1 answers
Gaurav Seth 5 years, 2 months ago
The owner of a taxi company decides to run all the taxis on CNG fuel instead of petrol/diesel. The taxi charges in city comprises of fixed charges together with the charge for the distance covered. For a journey of 12 km, the charge paid is 789 and for journey of 20 km, the charge paid is ₹145.
What will a person have to pay for travelling a distance of 30 km? (2014)
Solution:
Let the fixed charges = 7x
and the charge per km = ₹y
According to the Question,
Putting the value of y in (i), we get
x + 12(7) = 89
x + 84 = 89 ⇒ x = 89 – 84 = 5
Total fare for 30 km = x + 30y = 5 + 30(7)
= 5 + 210 = ₹215
Posted by Sameer Jaiswal 5 years, 2 months ago
- 1 answers
Gaurav Seth 5 years, 2 months ago
4. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of 
such that the sum of the numbers of the houses preceding the house numbered is equal to the sum of the numbers of the houses following it. Find this value of 
Ans. Here 
and 
= 
= 
= 
= 
= 
= 
= 49 x 25
According to question,
= 
+ = 
=
Since, is a counting number, so negative value will be neglected.
Posted by Chirag Sood 5 years, 2 months ago
- 1 answers
Yogita Ingle 5 years, 2 months ago
Since the mid point of PQ is the origin and PQ = 2a
OP = OQ = a
Hence the coordinates of P and Q are (0,a) and (0,-a) respectively.
Since triangles PQR and PQR' are equilateral
Their third vertices R and R' lie on the perpendicular bisector of base PQ.
X'OX is the perpendicular bisector of base PQ.
Thus R and R' lies on X-axis,
their Y coordinates are zero.
In 
ROP, OR2 + OP2 = PR2
OR2 + a2 = (2a)2
OR2 = 3a2
OR = 
Similarly, 
=
Thus,the coordinates of the vertices R and 
are 
respectively.
Posted by Aàyañ Âryäñ 5 years, 2 months ago
- 2 answers
Akhand Pratap Singh 5 years, 2 months ago
Yogita Ingle 5 years, 2 months ago
The bulb will glow because acid conduct electricity by releasing H+ ions in water and also a good conductor of electricity when HCL is replaced by NaOH the bulb still glows because base also conduct electricity by releasing OH- ions
Posted by Jagtaran Singh Maan 5 years, 2 months ago
- 0 answers
Posted by Puranjay Vyas 5 years, 2 months ago
- 1 answers
Yogita Ingle 5 years, 2 months ago
In triangle ABC given tan A is 4/3
hence opposite side is 4x, adjacent side is 3x
then by Pythagoras theorem hypotenuse will be 5x.
now,
sin A is 4/5
cos A is 3/5
cosec A is 5/4
sec A is 5/3
cot A is 3/4.
Posted by Ishrat Khan 5 years, 2 months ago
- 0 answers
Posted by Sonam Kumari Paswan 5 years, 2 months ago
- 3 answers
Pankaj Malik 5 years, 2 months ago
Posted by Dilraj Kâlêsh 5 years, 2 months ago
- 1 answers
Gaurav Seth 5 years, 2 months ago
Let us assume to the contrary that √3 is a rational number.
It can be expressed in the form of p/q
where p and q are co-primes and q≠ 0.
⇒ √3 = p/q
⇒ 3 = p2/q2 (Squaring on both the sides)
⇒ 3q2 = p2………………………………..(1)
It means that 3 divides p2 and also 3 divides p because each factor should appear two times for the square to exist.
So we have p = 3r
where r is some integer.
⇒ p2 = 9r2………………………………..(2)
from equation (1) and (2)
⇒ 3q2 = 9r2
⇒ q2 = 3r2
Where q2 is multiply of 3 and also q is multiple of 3.
Then p, q have a common factor of 3. This runs contrary to their being co-primes. Consequently, p / q is not a rational number. This demonstrates that √3 is an irrational number.
Posted by Simon Arulappan 5 years, 2 months ago
- 1 answers
Posted by Soumya Rajput 5 years, 2 months ago
- 0 answers
Posted by Kanagadurga V 5 years, 2 months ago
- 1 answers
Posted by Khush Kaur 5 years, 2 months ago
- 4 answers
Singh Krishna 5 years, 2 months ago
Posted by Nageena Kumari 5 years, 2 months ago
- 0 answers
Posted by Swathi.L ???? 5 years, 2 months ago
- 2 answers
Posted by Palak Jain 5 years, 2 months ago
- 0 answers
Posted by Palak Jain 5 years, 2 months ago
- 2 answers
Yogita Ingle 5 years, 2 months ago
(1) Given Equation is x + y = 7
x = 7 - y --------- (1)
(2) 2x - 3y = 11
2(7 - y) - 3y = 11
14 - 2y - 3y = 11
14 - 5y = 11
-5y = 11 - 14
-5y = -3
y =
Substitute y = in (1), we get
x = 7 -
=
=
Therefore the value of x = , y =
Posted by Palak Jain 5 years, 2 months ago
- 2 answers
Yogita Ingle 5 years, 2 months ago
Given pair of linear equations is
2x + 3y = 46 …(i)
And 3x + 5y = 74 …(ii)
On multiplying Eq. (i) by 3 and Eq. (ii) by 2 to make the coefficients of x equal, we get the equation as
6x + 9y = 138 …(iii)
6x + 10y = 148 …(iv)
On subtracting Eq. (iii) from Eq. (iv), we get
6x + 10y – 6x – 9y = 148 – 138
⇒ y = 10
On putting y = 10 in Eq. (ii), we get
3x + 5y = 74
⇒ 3x + 5(10) = 74
⇒ 3x + 50 = 74
⇒ 3x = 74 – 50
⇒ 3x = 24
⇒ x = 8
Hence, x = 8 and y = 10 , which is the required solution.
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Sarthak Gupta 5 years, 2 months ago
3Thank You