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=1,y=0
Step-by-step explanation:
X-2y=1.....equation 1
2x-y=2.....equation 2
X=1+2y.......equation 3
Substitute equation 3into 2
2(1+2y)-y=2
2+4y-y=2
2+3y=2
3y=2-2
3y=0
Y=0/3
Y=0
Substitute for y=0 in equation 1
X-2y=1
X-2(0)=1
X-0=1
X=1
X=1 & Y=0
Answer:
a= 6
Step-by-step explanation:
slope = (y2 - y1)/(x2 - x1)
3/8 = (3 - 0)/(a - (-2))
cross multiplying,
3(a + 2) = 3(8)
3a + 6 = 24
3a = 18
a = 18/3
a = 6
Answer:
BC = 11
Step-by-step explanation:
using the sine ratio in the right triangle and the exact value
sin45° = 
, then
sin 45° = 
= 
= 
= ( cross- multiply )
BC = 11
The correct question is
<span>In a circle with a radius of 3 ft, an arc is intercepted by a central angle of 2π/3 radians. What is the length of the arc?
we know that
in a circle
</span>2π radians -----------------> lenght of (2*π*r)
2π/3 radians--------------> X
X=[(2π/3)*(2π*r)]/[2π]=(2π/3)*r
the lenght of the arc=(2π/3)*3=2π ft
the answer is 2π ft
