$a=\dfrac{2d}{t^2}\ \ \ \ \ |\text{solve for t} \\ \\\\\dfrac{2d}{t^2}=\dfrac{a}{1}\ \ \ \ \ |\text{cross multiply}\\\\at^2=2d\ \ \ \ \ |:a\\\\t^2=\dfrac{2d}{a}\to t=\sqrt{\dfrac{2d}{a}}\\\\\text{We have}\\ \\a=2\dfrac{m}{s^2},\ d=10m\\\\\text{substitute}\\\\t=\sqrt{\dfrac{2\cdot10m}{2\frac{m}{s^2}}}=\sqrt{10m\cdot\dfrac{s^2}{m}}=\sqrt{10s^2}=\sqrt{10}\ s\\\\Answer:\ C.\ \sqrt{10}\ seconds$

4 0

3 years ago

A computer program found that the line y=3x - 52 is the best line of best fit to represent the temperature outside and the amoun

Aliun [14]

Answer:

Step-by-step explanation:

given x = 50

=> y = 3(50)-52 = 150-52=98

the ice cream they can expect to sell = 98

4 0

3 years ago