Answer:
2
Step-by-step explanation:
she will have 2 arangments of one rose and two daisies.
Answer:
Step-by-step explanation:
1. A car requires 22 litres of petrol to travel a distance of 259.6 km
what is the distance that the car can travel on 63 ltr of petrol
22ltr = 259.6km
63ltr=
cross multiply
{63 x 259.6}/22 = 16354.8/22 = 743.4 km
A car requires 22 litres of petrol to travel a distance of 259.6 km, it would require 63 ltr of petrol to travel 743.4km
2. To travel a distance of 2013.2 km
we would need to calculate the amount of fuel
A car requires 22 litres of petrol to travel a distance of 259.6 km
what amount of fuel would it require to travel 2013.2km
22ltr = 259.6km
xltr = 2013.2km
x is the value of petrol to cover 2013.2km
cross multiply
(2013.2 x 22)/259.6
44290.4/259.6 = 170.610169492≈170.6 ltr
A car requires 22 litres of petrol to travel a distance of 259.6 km, it would require 170.6 ltr of petrol to travel 2013.2km
if 1ltr is $1.99
170.6 ltr is (170.6 x 1.99)/1 = $339.494≈$339.5
The price of fuel consumed for 2013.2 km at 1 liter of petrol at $1.99 is $339.5
For this case we have:
Be a function of the form 
Where:
If we want to find f (-2), we substitute 
, then:
Since we have a negative root, the result will be given by complex numbers. By definition:
So:
Answer:
A) The longest horizontal distance is reached at 45 degrees angle. This is true for any projectile launch.
B) First, calculate fligth time (using the vertical motion) and then calculate the horizontal movement.
Flight time = 2* ascendent time
ascendent time => final vertical velocity, Vy, = 0
sin(45) = Voy / Vo => Voy = Vosin(45) = 25.5 m/s * (√2) / 2 = 18.03 m/s
Vy = Voy - gt = 0 => Voy = gt = t = Voy / g
Use g = 10 m/s^ (it is an aproximation, because the actual value is about 9.81 m/s^2 depending on the latitud)
t = 18.03 m/s / 10 m/s^2 = 1.83 s
This is the ascendant time going upward.
The flight time is 2*1.83 = 3.66 s
Horizontal motion
Horizontal velocity = Vx = constant = Vox = Vo*cos(45) = 18.03 m/s
Vx = x / t => x = Vx*t
Horizontal distance = xmax = 18.03m/s*3.66 s = 65.99 m
c) The time the ballon will be in the air was calculated in the part B, it is 18.03 s
