a=2, b=3a+=++a + ++b*b++-++awhat is the output​

Answer:

Data validation means checking the accuracy and quality of source data before using, importing or otherwise processing data. Different types of validation can be performed depending on destination constraints or objectives. Data validation is a form of data cleansing

Explanation:

Answer:

Data validation means checking the accuracy and quality of source data before using, importing or otherwise processing data. Different types of validation can be performed depending on destination constraints or objectives. Data validation is a form of data cleansing.

Explanation:

Answer:

The verb in the sentence is "affected".

• गृह — घर, निकेतन, भवन, आलय,निवास, गेह, सदन, आगार, आयतन, आवास, निलय, धाम।

• गंगा — सुरसरि, त्रिपथगा, देवनदी, जाह्नवी, भागीरथी।

• चंद्र — चाँद, चंद्रमा, विधु, शशि, राकेश, हिमांशु

Answer:

दो-दो समानार्थी शब्द लिखो- अद्भुत प्रशंसा देश

Explanation:

अतिथि- मेहमान, अभ्यागत, आगन्तुक, पाहुना।

अलंकार- आभूषण, भूषण, विभूषण, गहना, जेवर।

अहंकार- दंभ, गर्व, अभिमान, दर्प, मद, घमंड, मान।

अहंकारी- गर्वित, अकडू, मगरूर, अकड़बाज, गर्वीला, आत्माभिमानी, ठस्सेबाज, घमंडी।

अनुपम- अपूर्व, अतुल, अनोखा, अनूठा, अद्वितीय, अदभुत, अनन्य।

अश्व- हय, तुरंग, वाजि, घोडा, घोटक।

Step-by-step explanation:

Let the hundredth, tenth and ones digit of the given number of x, y, and z

Then the number will be

100x + 10y + z

The number is 17 times the sum of digits

⇒ 100x + 10y + z = 17 (x + y + z) = 17x + 17y + 17

⇒ 100x – 17x + 10y – 17y + z – 17z = 0

⇒ 83x – 7y – 16z = 0 ... (1)

Also if 198 is added to the number the digits sets reversed.

⇒ 100x + 10y + z + 198 = 100z + 10y + x

⇒ 100x – x + 10y – 10y + z – 100z + 198 = 0

⇒ 99x – 99z + 198 = 0

⇒ x – z + 2 = 0

⇒ z = x + 2 ... (2)

The the sum of the extreme digits is less than the middle digit by unity.

⇒ x + z = y – 1

= x – y + z + 1 = 0 ... (3)

Putting the value of z from (2) in (1) we get

83x – 7y – 16 (x + 2) = 0

⇒ 83x – 7y – 16x – 32 = 0

⇒ 67x – 7y = 32 ... (4)

Putting the value of z from (2) in (3) we get

x – y + x + 2 + 1 = 0

⇒ 2x – y + 3 = 0

⇒ y = 2x + 3 ... (5)

Putting the value of y from (5) in (4) we get

67x – 7 (2x + 3) = 32

67x– 14x – 21 = 32

53x = 32 + 21 = 53

⇒ x = 1 ... (6)

Putting the value of x from (6) in (8) we get

y = 2 + 3 = 5 ... (7)

Putting the value of x and y from (6) and (7) in (3) we get

1 – 5 + z + 1 = 0

⇒ z = 3

Hence the required number is 153.