When solving the problem 45x−3=16x+6 , your friend does the following: 45x−3=4(x+6) 5x−3=x+6 4x=9 x=94 Was this correct? (1 point) Responses No, the bases of 4 and 16 should have both been changed to a base of 2, giving us 22 and 24 respectively. No, the bases of 4 and 16 should have both been changed to a base of 2, giving us 2 squared and 2 to the 4th power respectively. No, in step 2, the 4 should have been distributed to the current exponents. This would change the equation and yield a different final answer. No, in step 2, the 4 should have been distributed to the current exponents. This would change the equation and yield a different final answer. Yes, this is correct. No error was made. Yes, this is correct. No error was made. No, in step 1 when the 16 changed to 4, it should have been 42 and the 2 distributed to the current exponent. No, in step 1 when the 16 changed to 4, it should have been 4 squared and the 2 distributed to the current exponent.

Question

When solving the problem 45x−3=16x+6 , your friend does the following: 45x−3=4(x+6) 5x−3=x+6 4x=9 x=94 Was this correct? (1 point) Responses No, the bases of 4 and 16 should have both been changed to a base of 2, giving us 22 and 24 respectively. No, the bases of 4 and 16 should have both been changed to a base of 2, giving us 2 squared and 2 to the 4th power respectively. No, in step 2, the 4 should have been distributed to the current exponents. This would change the equation and yield a different final answer. No, in step 2, the 4 should have been distributed to the current exponents. This would change the equation and yield a different final answer. Yes, this is correct. No error was made. Yes, this is correct. No error was made. No, in step 1 when the 16 changed to 4, it should have been 42 and the 2 distributed to the current exponent. No, in step 1 when the 16 changed to 4, it should have been 4 squared and the 2 distributed to the current exponent.

Answer 1