36=2*2*3*3=2²*3²
60=2*2*3*5=2²*3*5
LCM(36,60)=2*2*3*3*5=2²*3²*5=180
Answer:
Place one at (-7,-4) and the other at (-4,-6)
Step-by-step explanation:
Answer:
The launch angle should be adjusted to 30.63°
Step-by-step explanation:
The range of a projectile which is the horizontal distance covered by the projectile can be expressed as;
R =(v^2 sin2θ)/g
Where
R = range
v = initial speed
θ = launch angle
g = acceleration due to gravity
For the case above. When the projectile is launched at angle 13° above the horizontal.
θ1 = 13
R1 = (v^2 sin2θ1)/g
R1 = (v^2 sin26°)/g ....1
For the range to double
R2 = (v^2 sin2θ)/g .....2
R2 = 2R1
Substituting R2 and R1
(v^2 sin2θ)/g = 2 × (v^2 sin26°)/g
Divide both sides by v^2/g
sin2θ = 2sin26
2θ = sininverse(2sin26)
θ = sininverse(2sin26)/2
θ = 30.63°
Answer:
<em>Hi there!</em>
<u><em>For 1. it should be answer b</em></u>
<u><em>For 2. it would be answer c</em></u>
<u><em>For 3. it is answer a</em></u>
<em>Hope I Helped!</em>
