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:
a 1/ (6^2)
Step-by-step explanation:
6^3 / 6^6
when dividing exponents we can subtract
6^ (3-6)
6^(-3)
negative exponents go from the numerator to the denominator
1/ (6^3) = 1/216 = 6^-3
1/6^2 is not equal
Answer:
Area of parallelogram is 150 square ft
Step-by-step explanation:
Given the parallelogram in the figure whose height is 10 ft and base length is 15 ft.
Area of parallelogram is 
= 
= 150 square ft
now we have to decompose this parallelogram into rectangles and triangles and then to determine the shape.
The given parallelogram KLMN can be decomposed into 2 triangles and one rectangle.
2 right angled triangles KPL and NOM by construction

⇒
(∵ KLMN is a parallelogram)
Hence, dimension of triangles are 10 ft, 2 ft and 10.2 ft
Now rectangle KPON with dimension 15 ft and 10 ft because KP and NO are perpendiculars on LM.