by 14 1 4 each week for a number of weeks, w . He writes the expression 20(14) 20 ( 1 4 ) w to find the amount of radioactive su
1 answer:
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The complete question in the attached figure
step 1
find the tree height for an angle of elevation of 55°<span>
tan 55</span>°=opposite side angle 55°/adjacent side angle 55°
opposite side angle 55°=the tree height h1
adjacent side angle 55°=10 ft
so
tan 55°=h/10-------> h=10*tan 55°-----> h=14.28 ft
step 2
find the tree height for an angle of elevation of 75°
tan 75°=opposite side angle 75°/adjacent side angle 75°
opposite side angle 75°=the tree height h2
adjacent side angle 75°=10 ft
so
tan 75°=h2/10-------> h2=10*tan 75°-----> h2=37.32 ft
therefore
<span>the height of the tree is within the range [14.26,37.32]
</span>
Part 1) 4.2 ft
<span>the value is not in the interval [14.26,37.32]
therefore
it is </span>Not Reasonable
Part 2) 14.7 ft
the value is in the interval [14.26,37.32]
therefore
it is Reasonable
Part 3) 24.4 ft
the value is in the interval [14.26,37.32]
therefore
it is Reasonable
Part 4) 33.9 ft
the value is in the interval [14.26,37.32]
therefore
it is Reasonable
Part 5) 39.1ft
the value is not in the interval [14.26,37.32]
therefore
it is Not Reasonable
Part 6) 58.7 ft
the value is not in the interval [14.26,37.32]
therefore
it is Not Reasonable
step 1
find the tree height for an angle of elevation of 55°<span>
tan 55</span>°=opposite side angle 55°/adjacent side angle 55°
opposite side angle 55°=the tree height h1
adjacent side angle 55°=10 ft
so
tan 55°=h/10-------> h=10*tan 55°-----> h=14.28 ft
step 2
find the tree height for an angle of elevation of 75°
tan 75°=opposite side angle 75°/adjacent side angle 75°
opposite side angle 75°=the tree height h2
adjacent side angle 75°=10 ft
so
tan 75°=h2/10-------> h2=10*tan 75°-----> h2=37.32 ft
therefore
<span>the height of the tree is within the range [14.26,37.32]
</span>
Part 1) 4.2 ft
<span>the value is not in the interval [14.26,37.32]
therefore
it is </span>Not Reasonable
Part 2) 14.7 ft
the value is in the interval [14.26,37.32]
therefore
it is Reasonable
Part 3) 24.4 ft
the value is in the interval [14.26,37.32]
therefore
it is Reasonable
Part 4) 33.9 ft
the value is in the interval [14.26,37.32]
therefore
it is Reasonable
Part 5) 39.1ft
the value is not in the interval [14.26,37.32]
therefore
it is Not Reasonable
Part 6) 58.7 ft
the value is not in the interval [14.26,37.32]
therefore
it is Not Reasonable