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Sagot :
Adam can see the horizon at a distance of [tex]\sqrt{6}mi.[/tex]
Given that the height of Pam's eye level above sea level is [tex]324ft[/tex] and the height of adam's eye level above sea level is [tex]400ft[/tex].
The formula used to find the distance they see in miles is [tex]d=\sqrt{\frac{3h}{2}}[/tex].
Let the height of Pam's is [tex]h[/tex] and the height of adam's is [tex]h'[/tex].
The distance of Pam from the horizon can be given as [tex]d=\sqrt{\frac{3h}{2}}[/tex].
Substitute these values,
[tex]\begin{aligned}d&=\sqrt{3\times \frac{324}{2}}\\ d&=\sqrt{486}\\ d&=\sqrt{81\times 6}\\ d&=9\sqrt{6}\end[/tex]
And the distance of adam from the horizon can be given as [tex]d=\sqrt{\frac{3h'}{2}}[/tex].
Substitute these values,
[tex]d'=\sqrt{3\times \frac{400}{2}}\\ d'=\sqrt{600}\\ d'=\sqrt{100\times 6}\\ d'&=10\sqrt{6}[/tex]
The net distance of adam from the horizon is
[tex]\begin{aligned}d_{\text{net distance}}&=d'-d\\ &=10\sqrt{6}-9\sqrt{6}\\&=\sqrt{6}\end[/tex]
Hence, the adam can see the horizon at a distance of [tex]\sqrt{6}mi[/tex].
Learn more about horizon and distance here brainly.com/question/10016098
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