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:
16
Step-by-step explanation:
Answer:
you get 6.90372 so i thnk it is clse to B
SA: 2pi (r^2)+2pi(r)(h)
In order to find radius (r) of the base you must divide 10 by pi then find the square root of the result
√(10/pi)=r: 1.784cm
h: a) 8.0cm
b) 6.5cm
c) 9.4cm
2pi (3.183)+2pi(1.784)(8)=109.673cm^2
2pi (3.183)+2pi(1.784)(6.5)=92.859cm^2
2pi (3.183)+2pi(1.784)(9.4)=125.366cm^2
Answer:
No solutions
Step-by-step explanation:
i did this via substitution so i hope it isn't a problem
3x+y=4
6x+2y=−4
Step: Solve 3x+y=4for y:
3x+y+−3x=4+−3x(Add -3x to both sides)
y=−3x+4
Step: Substitute−3x+4foryin6x+2y=−4:
6x+2y=−4
6x+2(−3x+4)=−4
8=−4(Simplify both sides of the equation)
8+−8=−4+−8(Add -8 to both sides)
0=−12
so there are no solutions
hope this helps!!
