3 years ago

For how many integers $n$ is it true that $\sqrt{n} \le \sqrt{4n - 6} < \sqrt{2n + 5}$?

1 answer:

6 0

$\bf \sqrt{n}\leqslant \sqrt{4n-6}$

$\bf \sqrt{n}< \sqrt{2n+5}\implies \stackrel{\textit{squaring both sides}}{n< 2n+5}\implies 0\leqslant 2n - n + 5 \\\\\\ 0 < n+5\implies \boxed{-5 < n} \\\\\\ \stackrel{-5\leqslant n < 2}{\boxed{-5}\rule[0.35em]{10em}{0.25pt}0\rule[0.35em]{3em}{0.25pt}2}$

namely, -5, -4, -3, -2, -1, 0, 1. Excluding "2" because n < 2.

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A certain freely falling object, released from rest, requires 1.50 s to travel the last 30.0 m before it hit the ground. a.Find

Lyrx [107]

A.)
s= 30m
u = ? ( initial velocity of the object )
a = 9.81 m/s^2 ( accn of free fall )
t = 1.5 s
s = ut + 1/2 at^2
\[u = \frac{ S - 1/2 a t^2 }{ t }\]
\[u = \frac{ 30 - ( 0.5 \times 9.81 \times 1.5^2) }{ 1.5 } \]
\[u = 12.6 m/s\]

b.)
s = ut + 1/2 a t^2
u = 0 ,
s = 1/2 a t^2
\[s = \frac{ 1 }{ 2 } \times a \times t ^{2}\]
\[s = \frac{ 1 }{ 2 } \times 9.81 \times \left( \frac{ 12.6 }{ 9.81 } \right)^{2}\]
\[s = 8.0917...\]
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4 0

3 years ago

Can anyone answer this question please 6[y-2]/7-12=2[y-7]/3

Kaylis [27]

Answer:

y =47.5.

Step-by-step explanation:

First eliminate the fractions by multiplying through by the LCM of 7 and 3 which is 21:

21* 6[y-2]/7-21*12 = 21*2[y-7]/3

18(y - 2) - 252 = 14(y - 7)

18y -36 - 252 = 14y - 98

18y - 14y = -98 + 36 + 252