answer of ace with aster 7 practice book ch-10​

Answer:

It was ascended by silver steps and stood on golden feet set with jewels, and it was backed by representations of two open peacocks' tails, gilded, enamelled, and inset with diamonds, rubies, and other stones. The throne was seized along with other plunder when the Iranian conqueror Nādir Shāh captured Delhi in 1739.

Explanation:

Answer:

It was ascended by silver steps and stood on golden feet set with jewels, and it was backed by representations of two open peacocks' tails, gilded, enamelled, and inset with diamonds, rubies, and other stones. The throne was seized along with other plunder when the Iranian conqueror Nādir Shāh captured Delhi in 1739.

Explanation:

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Answer:

b is write or not this is write or not

Answer:

The correct answer will be -

लेन्सची शक्ती +4.0 D असेल. The power of the lens will be +4.0 D.

Explanation:

Given in the question -

• बहिर्वक्र भिंगाची फोकल लांबी 25 सेमी आहे.

आता आपल्याला लेन्सची शक्ती शोधावी लागेल.

आपल्याला माहित आहे की लेन्सची शक्ती ही लेन्सच्या फोकल लांबीची परस्पर आहे.We know that power of the lens is the reciprocal of the focal length of the lens.

So, power of the lens =  [tex]\frac{1}{focal\ length(in \ m)}[/tex]

Focal length = +25 cm = +0.25 m

Power of the lens = [tex]\frac{1}{0.25} = + 4.0\ D[/tex]

So,लेन्सची शक्ती +4.0 D असेल (the power of the lens will be +4.0 D).

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Answer:

अपना काम करने से

Explanation:

अपना काम करने से हम भगवान को प॔पत कर सकते हैं ।

Answer:

c

Explanation:

C IS THE CORRECT ANSWER

Answer:

Identity I: (a + b)2 = a2 + 2ab + b2

Identity II: (a – b)2 = a2 – 2ab + b2

Identity III: a2 – b2= (a + b)(a – b)

Identity IV: (x + a)(x + b) = x2 + (a + b) x + ab

Identity V: (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

Identity VI: (a + b)3 = a3 + b3 + 3ab (a + b)

Identity VII: (a – b)3 = a3 – b3 – 3ab (a – b)

Identity VIII: a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)

Step-by-step explanation:

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[tex]\large\mathfrak\pink{♡answer♡}[/tex]

• Identity I: (a + b)² = a² + 2ab + b²
• Identity II: (a – b)² = a² – 2ab + b²
• Identity III: a² – b² = (a + b)(a – b)
• Identity IV: (x + a)(x + b) = x² + (a + b) x +ab
• Identity V: (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
• Identity VI: (a + b)3 = a³ + b³ + 3ab (a + b)
• Identity VII: (a – b)³ = a³ – b³ – 3ab (a – b)
• Identity VIII: a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca)