The original time would be 37 minutes before 6:30 so you would subtract 37 minutes from 6;30. So 6:30 minus 30 would be 6 and you would have 7 left over so 6:00 minus 7 minutes would be 5:53

6 0

3 years ago

Solve the equation by completing the square.Round to the nearest hundredth if necessary. X^2–x–7=0

lesya [120]

NOTE: $\left(x-\dfrac{1}{2}\right)^2=x^2-x+\dfrac{1}{4}$

$x^2-x-7=0 \\ \\ \underbrace{x^2-x+\frac{1}{4}}_{\left(x-\frac{1}{2}\right)^2}-7\dfrac{1}{4}=0 \\ \\ \\ \left(x-\frac{1}{2}\right)^2=7\frac{1}{4}=\frac{29}{4} \qquad /\sqrt{}\\ \\ \left|x-\frac{1}{2}\right|=\dfrac{\sqrt{29}}{2} \\ \\ x_1=\dfrac{1}{2}-\dfrac{\sqrt{29}}{2} \approx -2.19 \\ \\ x_2=\dfrac{1}{2}+\dfrac{\sqrt{29}}{2} \approx 3.19$

ANSWER: c)

5 0

3 years ago