Answer:
B. I got it right on the quiz
Step-by-step explanation:
For the first question:Using the distance=rate x time formula (d=rt). Your equation would be y=38x. The 38 comes from the kathryn’s speed(rate).
For the second question: because you know the distance is now 53 substitute 53 for y and solve. 53=38x. Where x represents her time. She travels the 53 miles at a rate of 38mi/h (53 divided by 38) and solve in terms of hours 1.3947or rounded to the nearest tenth 1.4 which in terms of hours is 1 hour 24 minutes.
829 is rounded to 830 because the 9 is greater than 4 so its going to be rounded up.
Given quadratic equation
![](https://tex.z-dn.net/?f=)
![x^2+6x+6=0](https://tex.z-dn.net/?f=x%5E2%2B6x%2B6%3D0)
Using the usual conventions,
a=1, b=6, c=6
Apply the quadratic formula,
![x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-b%5Cpm%5Csqrt%7Bb%5E2-4ac%7D%7D%7B2a%7D)
substitute a,b & c with numerical values,
![x=\frac{-6\pm\sqrt{6^2-4(1)(6)}}{2(1)}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-6%5Cpm%5Csqrt%7B6%5E2-4%281%29%286%29%7D%7D%7B2%281%29%7D)
simplify:
in log minus means divide
log 6 - log 3 = log (6÷3)
log 6 - log 3 = log 2