Hello Austint1414
The answer in standard form would be <u><em>70,633,999,564,087,001</em></u>
:)
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>
Answer:
Ic² + b²l = 13 units.
Step-by-step explanation:
We have to evaluate the expression Ic² + b²l with unknowns b and c and having the values of b and c respectively - 3 and - 2.
Now, Ic² + b²l
= I(- 2)² + (- 3)²l {Putting the values of b and c}
= I4 + 9l
= I13l
= 13 units.
Therefore, Ic² + b²l = 13 units. (Answer)
