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Ask QuestionPosted by Jayank Mishra 5 years, 1 month ago
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Posted by Om Shukla 5 years, 1 month ago
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Parul Ailawadi 5 years, 1 month ago
Anuj Mathpal 5 years, 1 month ago
Posted by Ryan George 5 years, 1 month ago
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Gaurav Seth 5 years ago
Let AB be a diameter of the circle. Two tangents PQ and RS are drawn at points A and B respectively.
Radius drawn to these tangents will be perpendicular to the tangents.
Thus, OA ⊥ RS and OB ⊥ PQ
∠OAR = 90º
∠OAS = 90º
∠OBP = 90º
∠OBQ = 90º
It can be observed that
∠OAR = ∠OBQ (Alternate interior angles)
∠OAS = ∠OBP (Alternate interior angles)
Since alternate interior angles are equal, lines PQ and RS will be parallel.
Posted by Md Shahid Afridi 5 years, 1 month ago
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Sourav Keshri 5 years, 1 month ago
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Posted by Harsh Gupta X 'C' 5 years, 1 month ago
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Gaurav Seth 5 years ago
Given,
Power of lens, P = - 2 D
Therefore,
Focal length,
f = 1 / P
= 1 / 2 m = - 0.5
The negative sign of focal length indicates that the lens is concave.
Given f = - 20 cm, (sign convention) , v = - 10 cm
(∴ image formed by concave lens is virtual)
By using lens formula,
or u = - 20
Thus the object is placed at 20 cm from the concave lens
Posted by Aarohi Tiwari 5 years, 1 month ago
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Gaurav Seth 5 years, 1 month ago
The equation whose graph is a straight line is called linear equation.
Linear equations are also first-degree equations as it has the highest exponent of variables as 1. Some of the examples of such equations are as follows:
- 2x – 3 = 0,
- 2y = 8
- m + 1 = 0,
- x/2 = 3
Posted by Vishnu Singh 5 years, 1 month ago
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Gaurav Seth 5 years, 1 month ago
Both assumed mean method and step deviation method are ways to find mean without any complications...
step deviation method is just an elaboration of assumed mean.
1. Assumed Mean method:
Step 1- Tabulate the continuous class intervals and frequency alongwith class marks.
Class Mark = Upper Limit + Lower Limit
2
Step 2- Assume any of the class marks to be your mean .
Step 3- Take out the differences of the assumed mean and class marks.{ xi - a = di}
Example if the class marks are 20, 30,40,50,60,70 and so on......and if the a =50 then,di is -30,-20,-10,0,10,20and so on.....
Step 4: Multiply fi and di .
Step5:
d = sum of fidi
sum of fi
Step 6: Mean = a + d
Posted by Kritika Pathak 5 years, 1 month ago
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Posted by Aditya Singh 5 years, 1 month ago
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Gaurav Seth 5 years, 1 month ago
Q : The length of the hypotenuse of a right triangle exceeds the length of its base by 2 cm and exceeds twice the length of altitude by 1 cm. Find the length of each side of the triangle.
S o l u t i o n:
Let altitude of triangle be x.
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Posted by Tanushree Bodkhe 5 years, 1 month ago
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Gaurav Seth 5 years ago
TEP DEVIATION METHOD:
Step deviation method is used in the cases where the deviation from the assumed mean 'A' are multiples of a common number. If the values of ‘di’ for each class is a multiple of ‘h’ the calculation become simpler by taking ui= di/h = (xi - A )/h
Here, h is the class size of each class interval.
★★ Find the class marks of class interval. These class marks would serve as the representative of whole class and are represented by xi.
★★ Class marks (xi) = ( lower limit + upper limit) /2
★★ We may take Assumed mean 'A’ to be that xi which lies in the middle of x1 ,x2 …..xn
MEAN = A + h ×(Σfiui /Σfi) , where ui = (xi - A )/h
[‘Σ’ Sigma means ‘summation’ ]
FREQUENCY DISTRIBUTION TABLE
★★ Here,the frequency table is given in inclusive form. So we first change it into exclusive form by subtracting and adding h/2 to the lower and upper limits respectively of each class ,where ‘h’ denotes the difference of lower limit of a class and upper limit of the previous class.
Here, h/2 = ½ = 0.5
From the table : Σfiui = 25 , Σfi = 400
Let the assumed mean, A = 57, h = 3
MEAN = A + h ×(Σfiui /Σfi)
Mean = 57 + 3 (25/400)
= 57 + 3 (1/16)
= 57 + 3/16
= 57 + 0.187
= 57.187
Mean = 57.19 (approximate)
Hence, the Mean number of mangoes kept in a packing box is 57.19 (approximate)
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Rîtíkã Çhäûhāñ 5 years, 1 month ago
Gaurav Seth 5 years, 1 month ago
Answer:
g(x) = x² - x + 1
Step-by-step explanation:
On dividing x3-3x2+x+2 by a polynomial g(x),the quotient and remainder were x-2 and -2x+4 respectively.find g(x)
f(x) = x³ - 3x² + x + 2
q(x) = x - 2
let say
g(x) = ax² + bx + c
r = - 2x + 4
f(x) = g(x)q(x) + r
=> x³ - 3x² + x + 2 = (ax² + bx + c)(x - 2) + (-2x + 4)
=> x³ - 3x² + x + 2 = ax³ + x²(b - 2a) + x(c - 2b) -2c -2x + 4
=> x³ - 3x² + x + 2 = ax³ + x²(b - 2a) + x(c - 2b - 2) + (-2c + 4)
Equating Power terms
a = 1
b - 2a = - 3 => b -2 = -3 => b = -1
c - 2b - 2 = 1 => c + 2 - 2 = 1 => c = 1
or -2c + 4 = 2 => -2c = -2 => c = 1
g(x) = x² - x + 1
Posted by Jot Ughra 5 years, 1 month ago
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Rîtíkã Çhäûhāñ 5 years, 1 month ago
Posted by Mohit Parmar 5 years, 1 month ago
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Posted by Tanveer Kaur 5 years, 1 month ago
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Yogita Ingle 5 years, 1 month ago
The given sequence is an A.P. with first term a=3 and common difference d=3. Let there be n terms in the given sequence. Then,
nth term=111
a+(n−1)d=111
3+(n−1)×3=111
n=37
Thus, the given sequence contains 37 terms.
Posted by Siddhodhan Maale 5 years, 1 month ago
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Rani Kumari 5 years, 1 month ago
Yogita Ingle 5 years, 1 month ago
» 2x + 3y + 5 = 0
» kx + 4y = 10
➡ kx + 4y - 10 = 0
we have to find the value of k such that the pair of equations have unique solution.
we know that, when pair of equations have unique solution,
a1/a2 ≠ b1/b2
➡ 2/k ≠ 3/5
just put any value of k so that it doesn't get equal to 3/5.
you can put 1, 2 etc.
for example, k is 2, then a1/a2 = 2/2 = 1
LHS ≠ RHS
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Posted by Tanveer Kaur 5 years, 1 month ago
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Yogita Ingle 5 years, 1 month ago
Let the three numbers in AP be a-d, a, a+d
Then,
sum = a-d+a+a+d=15
3a = 15
So, a = 5
Also, Sum of squares = 83
(a-d)2+a2+(a+d)2=83
(5-d)2+25+(5+d)2=83
25+d2-10d+25+25+d2+10d=83
75+2d2=83
2d2=8
d=+2 or -2
When d = 2,
The numbers are 3, 5, 7
When d = -2,
The numbers are 7, 5, 3
Posted by Harsh Singh 5 years, 1 month ago
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Yogita Ingle 5 years, 1 month ago
Let the selling price (S. P.) be x.
Hence, selling price is ₹ 476.
Posted by Venkataramana Shenoy 5 years, 1 month ago
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Yogita Ingle 5 years, 1 month ago
The first 8 multiples of 8 are
8, 16, 24, 32, 40, 48, 56,64
These are in an A.P., having first term as 8 and common difference as 8.
Therefore, a = 8
d = 8
S15 = ?
Sn = n/2 [2a + (n - 1)d]
S15 = 15/2 [2(8) + (15 - 1)8]
= 15/2[6 + (14) (8)]
= 15/2[16 + 112]
= 15(128)/2
= 15 × 64
= 960
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Gaurav Seth 5 years, 1 month ago
1. Find the roots of the following quadratic equations, if they exist, by the method of completing the square:
(i) 2x2 – 7x +3 = 0
(ii) 2x2 + x – 4 = 0
(iii) 4x2 + 4√3x + 3 = 0
(iv) 2x2 + x + 4 = 0
Solutions:
(i) 2x2 – 7x + 3 = 0
⇒ 2x2 – 7x = – 3
Dividing by 2 on both sides, we get
⇒ x2 -7x/2 = -3/2
⇒ x2 -2 × x ×7/4 = -3/2
On adding (7/4)2 to both sides of equation, we get
⇒ (x)2-2×x×7/4 +(7/4)2 = (7/4)2-3/2
⇒ (x-7/4)2 = (49/16) – (3/2)
⇒(x-7/4)2 = 25/16
⇒(x-7/4)2 = ±5/4
⇒ x = 7/4 ± 5/4
⇒ x = 7/4 + 5/4 or x = 7/4 – 5/4
⇒ x = 12/4 or x = 2/4
⇒ x = 3 or x = 1/2
(ii) 2x2 + x – 4 = 0
⇒ 2x2 + x = 4
Dividing both sides of the equation by 2, we get
⇒ x2 +x/2 = 2
Now on adding (1/4)2 to both sides of the equation, we get,
⇒ (x)2 + 2 × x × 1/4 + (1/4)2 = 2 + (1/4)2
⇒ (x + 1/4)2 = 33/16
⇒ x + 1/4 = ± √33/4
⇒ x = ± √33/4 – 1/4
⇒ x = ± √33-1/4
Therefore, either x = √33-1/4 or x = -√33-1/4
(iii) 4x2 + 4√3x + 3 = 0
Converting the equation into a2+2ab+b2 form, we get,
⇒ (2x)2 + 2 × 2x × √3 + (√3)2 = 0
⇒ (2x + √3)2 = 0
⇒ (2x + √3) = 0 and (2x + √3) = 0
Therefore, either x = -√3/2 or x = -√3/2.
(iv) 2x2 + x + 4 = 0
⇒ 2x2 + x = -4
Dividing both sides of the equation by 2, we get
⇒ x2 + 1/2x = 2
⇒ x2 + 2 × x × 1/4 = -2
By adding (1/4)2 to both sides of the equation, we get
⇒ (x)2 + 2 × x × 1/4 + (1/4)2 = (1/4)2 – 2
⇒ (x + 1/4)2 = 1/16 – 2
⇒ (x + 1/4)2 = -31/16
As we know, the square of numbers cannot be negative.
Therefore, there is no real root for the given equation, 2x2 + x + 4 = 0.
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