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Schach [20]
3 years ago
14

The function h(t)=-4.87t^2+18.75t is used to model the height of an object projected in the air, where h(t) is the height in met

ers and t is the time in seconds. Rounded to the nearest hundredth, what are the domain and range of the function h(t)?

Mathematics
1 answer:
Kazeer [188]3 years ago
4 0
The correct answer is the first choice.

For the domain, we are looking at the included x-values. The values go from 0 to about 3.85.

For the range, we are looking at the included y-values. The values go from 0 to the top height of about 18.05.
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\fbox{ \fbox { \sf{Distance  =  \sf{15 \: units}}}}

{ \fbox { \fbox { \sf{Midpoint = { \sf{ \: (12 \: , \: 10.5)}}}}}}

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\star{ \:  \sf { \: Let \: the \: points \: be \: A \: and \: B}}

\star { \sf{Let \: A(6 \:, 6) \: be \: (x1 ,\: y1) \: and \: B(18 ,\: 15) \: be \: (x2 \:, y2)}}

\underline{ \underline{ \tt{Finding \: the \: distance}}}

\boxed{ \sf{Distance =  \sqrt{ {(x2 - x1)}^{2}  +  {(y2 - y1)}^{2} } }}

\mapsto{ \sf{Distance =  \sqrt{ {(18 - 6)}^{2}  +  {(15 - 6)}^{2} } }}

\mapsto{ \sf{Distance =  \sqrt{ {(12)}^{2}  +  {(9)}^{2} } }}

\mapsto{ \sf{Distance =  \sqrt{144 + 81}}}

\mapsto{ \sf{Distance =  \sqrt{225} }}

\mapsto{ \sf{Distance =  \sqrt{ {15}^{2} } }}

\mapsto{  \boxed{\sf{Distance = 15 \: units}}}

\underline{ \underline {\tt{Finding \: the \: Midpoint}}}

\boxed{ \sf{Midpoint = ( \frac{x1 + x2}{2}  \: , \:  \frac{y1 + y2}{2} )}}

\mapsto{ \sf{Midpoint = ( \frac{6 + 18}{2}  \: , \:  \frac{6 + 15}{2} )}}

\mapsto{ \sf{Midpoint = ( \frac{24}{2}  \: , \:  \frac{21}{2} )}}

\mapsto{ \boxed{ \sf{Midpoint = (12 \: , \: 10.5)}}}

Hope I helped!

Best regards! :D

~\sf{TheAnimeGirl}

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