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:
x = (6 - y)÷ 3
Step-by-step explanation:
3x + y = 6
3x = 6 - y
x = 6 - y ÷ 3
The answer is 21 + 4x.
You find this answer by adding up all the sides.
9+12+4x= 21+4x ... you don't add 4x because it has a variable that we do not know of the value.
Answer:
1 × 3 + 4 - 2 = 5
Step-by-step explanation:
1 × 3 = 3
3 + 4 = 7
7 - 2 = 5
Answer:
11
Step-by-step explanation:
