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Ask QuestionPosted by Navya Sravani 4 years, 5 months ago
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Adithya Dev A 4 years, 5 months ago
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Preeti Dabral 4 years, 3 months ago
Given, height of object = 5cm
Position of object, u = - 25cm
Focal length of the lens, f = 10 cm
Hence, position of image, v =?
We know that,
1/v - 1/u = 1/f
1/v + 1/25 = 1/10
So, 1/v = 1/10 - 1/25
S0, 1/v = (5 - 2)/50
That is, 1/v = 3/50
So, v= 50/3 = 16.66 cm
Thus, distance of image is 16.66 cm on the opposite side of lens.
Now, magnification = v/u
That is, m = 16.66/-25 = -0.66
Also, m= height of image/height of object
Or, -0.66 = height of image / 5 cm
Therefore, height of image = -3.3 cm
The negative sign of height of image shows that an inverted image is formed.
Thus, position of image = At 16.66 cm on opposite side of lens
Size of image = - 3.3 cm at the opposite side of lens
Nature of image – Real and inverted....
show how would you connect the three resistors, each of resistance 6 so that the combination has a resistance of 9 , 4
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Pankaj Kumar Patel 4 years, 5 months ago
Rock ✌?✌? 4 years, 5 months ago
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Adithya Dev A 4 years, 5 months ago
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Sia ? 4 years, 5 months ago
11x + 15y = -23 ---- (1) ////// 7x - 2y = 20 ---- (2). Take equation (2). Thus, x = 20 + 2y / 7// ---- (A). Equating (A) to (1), we get :- 11 (20 + 2y / 7) + 15y = -23 => 220 + 22y / 7 + 15y = -23 => 220 + 22y + 105y = -161 => 127y = -161 - 220 => 127y = -381 => y = -3// Therefore, x = 20 + 2y / 7 = 20 - 6 / 7 = 14 / 7 = 2
Posted by S Harish 4 years, 5 months ago
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Sia ? 4 years, 5 months ago
- Select Insert > Online Pictures.
- Type a word or phrase to describe what you're looking for, then press Enter.
- Filter the results by Type for Clipart.
- Select a picture.
- Select Insert.
Posted by Snowie Bakshi 4 years, 3 months ago
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Preeti Dabral 4 years, 3 months ago
According to Euclid’s Division Lemma if we have two positive integers a and b, then there exist unique integers q and r which satisfies the condition a = bq + r where 0 ≤ r ≤ b.
HCF is the largest number which exactly divides two or more positive integers.
Since 12576 > 4052
12576 = (4052 × 3) + 420
420 is a reminder which is not equal to zero (420 ≠ 0).
4052 = (420 × 9) + 272
271 is a reminder which is not equal to zero (272 ≠ 0).
Now consider the new divisor 272 and the new remainder 148.
272 = (148 × 1) + 124
Now consider the new divisor 148 and the new remainder 124.
148 = (124 × 1) + 24
Now consider the new divisor 124 and the new remainder 24.
124 = (24 × 5) + 4
Now consider the new divisor 24 and the new remainder 4.
24 = (4 × 6) + 0
Reminder = 0
Divisor = 4
HCF of 12576 and 4052 = 4.
Posted by Prakash Jaiswal 4 years, 5 months ago
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Adithya Dev A 4 years, 5 months ago
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Adithya Dev A 4 years, 5 months ago
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Navya Sravani 4 years, 5 months ago
1Thank You