<h2><em>Answer:</em></h2><h2><em>Answer:x = 7+4√3</em></h2><h2 /><h2><em>Answer:x = 7+4√3To find √x we proceed,</em></h2><h2 /><h2><em>Answer:x = 7+4√3To find √x we proceed,√x = √(7+4√3)</em></h2><h2 /><h2><em>Answer:x = 7+4√3To find √x we proceed,√x = √(7+4√3)√x = √(7+2x2√3)</em></h2><h2 /><h2><em>Answer:x = 7+4√3To find √x we proceed,√x = √(7+4√3)√x = √(7+2x2√3)√x = √(7+2√3x4)</em></h2><h2 /><h2><em>Answer:x = 7+4√3To find √x we proceed,√x = √(7+4√3)√x = √(7+2x2√3)√x = √(7+2√3x4)√x = √(3+4+2√3x4)….. {writing 7 = 3+4}</em></h2><h2 /><h2><em>Answer:x = 7+4√3To find √x we proceed,√x = √(7+4√3)√x = √(7+2x2√3)√x = √(7+2√3x4)√x = √(3+4+2√3x4)….. {writing 7 = 3+4}If we observe RHS of √x we observe form of</em></h2><h2 /><h2><em>Answer:x = 7+4√3To find √x we proceed,√x = √(7+4√3)√x = √(7+2x2√3)√x = √(7+2√3x4)√x = √(3+4+2√3x4)….. {writing 7 = 3+4}If we observe RHS of √x we observe form of√(a² + b² +2ab) where a=√3 and b =√4</em></h2><h2 /><h2><em>Answer:x = 7+4√3To find √x we proceed,√x = √(7+4√3)√x = √(7+2x2√3)√x = √(7+2√3x4)√x = √(3+4+2√3x4)….. {writing 7 = 3+4}If we observe RHS of √x we observe form of√(a² + b² +2ab) where a=√3 and b =√4Hence, √x =√(√3 +√4)² = √3 + √4 = 2+√3</em></h2><h2 /><h2><em>Answer:x = 7+4√3To find √x we proceed,√x = √(7+4√3)√x = √(7+2x2√3)√x = √(7+2√3x4)√x = √(3+4+2√3x4)….. {writing 7 = 3+4}If we observe RHS of √x we observe form of√(a² + b² +2ab) where a=√3 and b =√4Hence, √x =√(√3 +√4)² = √3 + √4 = 2+√3√x = 2+√3</em></h2><h2 /><h2><em>Answer:x = 7+4√3To find √x we proceed,√x = √(7+4√3)√x = √(7+2x2√3)√x = √(7+2√3x4)√x = √(3+4+2√3x4)….. {writing 7 = 3+4}If we observe RHS of √x we observe form of√(a² + b² +2ab) where a=√3 and b =√4Hence, √x =√(√3 +√4)² = √3 + √4 = 2+√3√x = 2+√31/√x = 1/(2+√3)</em></h2><h2 /><h2><em>Answer:x = 7+4√3To find √x we proceed,√x = √(7+4√3)√x = √(7+2x2√3)√x = √(7+2√3x4)√x = √(3+4+2√3x4)….. {writing 7 = 3+4}If we observe RHS of √x we observe form of√(a² + b² +2ab) where a=√3 and b =√4Hence, √x =√(√3 +√4)² = √3 + √4 = 2+√3√x = 2+√31/√x = 1/(2+√3)Multiplying both numerator and denominator by 2 - √3, we get</em></h2><h2 /><h2><em>Answer:x = 7+4√3To find √x we proceed,√x = √(7+4√3)√x = √(7+2x2√3)√x = √(7+2√3x4)√x = √(3+4+2√3x4)….. {writing 7 = 3+4}If we observe RHS of √x we observe form of√(a² + b² +2ab) where a=√3 and b =√4Hence, √x =√(√3 +√4)² = √3 + √4 = 2+√3√x = 2+√31/√x = 1/(2+√3)Multiplying both numerator and denominator by 2 - √3, we get1/√x = (2-√3)/(2-√3)(2+√3) = (2-√3)/(2²-√3²) =</em></h2><h2 /><h2><em>Answer:x = 7+4√3To find √x we proceed,√x = √(7+4√3)√x = √(7+2x2√3)√x = √(7+2√3x4)√x = √(3+4+2√3x4)….. {writing 7 = 3+4}If we observe RHS of √x we observe form of√(a² + b² +2ab) where a=√3 and b =√4Hence, √x =√(√3 +√4)² = √3 + √4 = 2+√3√x = 2+√31/√x = 1/(2+√3)Multiplying both numerator and denominator by 2 - √3, we get1/√x = (2-√3)/(2-√3)(2+√3) = (2-√3)/(2²-√3²) =1/√x = 2-√3</em></h2><h2 /><h2><em>Answer:x = 7+4√3To find √x we proceed,√x = √(7+4√3)√x = √(7+2x2√3)√x = √(7+2√3x4)√x = √(3+4+2√3x4)….. {writing 7 = 3+4}If we observe RHS of √x we observe form of√(a² + b² +2ab) where a=√3 and b =√4Hence, √x =√(√3 +√4)² = √3 + √4 = 2+√3√x = 2+√31/√x = 1/(2+√3)Multiplying both numerator and denominator by 2 - √3, we get1/√x = (2-√3)/(2-√3)(2+√3) = (2-√3)/(2²-√3²) =1/√x = 2-√3Hence √x +1/√x = 2+√3 +2 -√3 = 4</em></h2><h2 />
In an Arithmetic sequence, the difference between two consecutive terms of the sequence remains the same. This difference is known as common difference.
If we observe the given options, we can note:
In option A, the difference between two consecutive terms is the same and is equal to 1.
In option B, the difference between two consecutive terms is the same and is equal to -2.
In option C, the difference is not the same.
In option D, the difference between two consecutive terms is the same and is equal to 5.
In option E, the difference is not the same.
So, the correct options are A,B and D
The domain is all real numbers and the range is all real numbers f(x) such that f(x) ≥ 7. C) The domain is all real numbers x such that x ≥ 0 and the range is all real numbers. Incorrect. Negative values can be used for x, but the range is restricted because x2 ≥ 0.
