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Whitepunk [10]
3 years ago
11

Please help. Evaluate 6! 3,125 720 120

Mathematics
1 answer:
Anastasy [175]3 years ago
4 0

\huge\textsf{Hey there!}

\large\textsf{6!}\\\large\textsf{= 6}\times\large\textsf{5}\times\large\textsf{4}\times\large\textsf{3}\times\large\textsf{2}\times\large\textsf{1}\\\large\textsf{6(5) = \bf 30}\\\large\textsf{= 30}\times\large\textsf{4}\times\large\textsf{3}\times\large\textsf{2}\times\large\textsf{1}\\\large\textsf{30(4) = \bf  120}\\\large\textsf{= 120}\times\large\textsf{3}\times\large\textsf{2}\times\textsf{1}\\\large\textsf{120(3) = \bf 360}\\\large\textsf{= 360}\times\large\textsf{2}\times\large\textsf{1}

\large\textsf{360(2) = \bf 720}\\\large\textsf{720}\times\large\textsf{1}\\\large\textsf{= \bf 720}

\boxed{\boxed{\huge\textsf{Therefore, your answer is: \bf 720}\huge\textsf{ (option B)}}}\huge\checkmark

\large\textsf{Good luck on your assignment and enjoy your day!}

~\frak{Amphitrite1040:)}

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marissa [1.9K]

Answer:

x = 28.01t,

y = 10.26t - 4.9t^2 + 2

Step-by-step explanation:

If we are given that an object is thrown with an initial velocity of say, v1 m / s at a height of h meters, at an angle of theta ( θ ), these parametric equations would be in the following format -

x = ( 30 cos 20° )( time ),

y = - 4.9t^2 + ( 30 cos 20° )( time ) + 2

To determine " ( 30 cos 20° )( time ) " you would do the following calculations -

( x = 30 * 0.93... = ( About ) 28.01t

This represents our horizontal distance, respectively the vertical distance should be the following -

y = 30 * 0.34 - 4.9t^2,

( y = ( About ) 10.26t - 4.9t^2 + 2

In other words, our solution should be,

x = 28.01t,

y = 10.26t - 4.9t^2 + 2

<u><em>These are are parametric equations</em></u>

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