<u>it</u><u> </u><u>is</u><u> </u><u>a</u><u> </u><u>type</u><u> </u><u>of</u><u> </u><u>trinomial</u><u> </u><u>polynomi</u><u>al</u><u> </u><u>because</u><u> </u><u>it contains </u><u>3</u><u> </u><u>terms</u><u> </u><u>with</u><u> </u><u>fundamen</u><u>tal</u><u> </u><u>signs</u><u>.</u><u> </u>
<u>hope</u><u> </u><u>it</u><u> </u><u>helps</u><u> </u><u>you</u>
Answer: C
Explanation: they will never touch
4a<52
All you have to do is divide 4 from a and 52.
4a<52
/4 /4
a<13
a<208 √
a<56 √
a<48 √
a<13 X
So the odd one out is a<13
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hope it helps
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°