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°
You have to multiply first and then be a bad bleep
Product rule- the first rule for simplifying square roots, which allows us to separate the square root of a product of two numbers into the product of two separate rational expressions
HOPES THIS HELP :)
We can write this as the difference of squares:
(5b⁸+8c)(5b⁸-8c)
To write as the difference of squares, take the square root of each term first:
√25b¹⁶ = 5b⁸; √64c² = 8c
Now we write this as a sum in one binomial and a difference in the other:
(5b⁸+8c)(5b⁸-8c)
Answer:
x=75
Step-by-step explanation:
The triangle is isosceles, meaning that the angles V and U are congruent
