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olga_2 [115]
2 years ago
11

Help me pleasehelp me please​

Mathematics
1 answer:
Nikolay [14]2 years ago
7 0

Answer:

1 second

Step-by-step explanation:

s=-5\\u=0\\v=\\a=-10\\t=t

\begin{aligned}\textsf{Using} \quad s & =ut+\dfrac12at^2\\ \implies -5 & =(0)t+\dfrac12(-10)t^2\\ -5 & = -5t^2\\ t^2& = 1\\ t & = \pm\sqrt{1}\\ t & = 1\: \sf s\end{aligned}

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First, let's see how 23 compares with the squares of the positive whole numbers on the number line.

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Look at the attached image to see where I plotted the approximate location of √23.

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