Answer:
0.25 rad to the nearest hundredth radian
Step-by-step explanation:
Here is the complete question
Suppose a projectile is fired from a cannon with velocity vo and angle of elevation (theta). The horizontal distance R(θ) it travels (in feet) is given by the following.
R(θ) = v₀²sin2θ/32
If vo=80ft/s what angel (theta) (in radians) should be used to hit a target on the ground 95 feet in front of the cannon?
Do not round any intermediate computations, and round your answer(s) to the nearest hundredth of a radian.
(θ)= ?rad
Solution
R(θ) = v₀²sin2θ/32
If v₀ = 80 ft/s and R(θ) = 95 ft
θ = [sin⁻¹(32R(θ)/v₀²)]/2
= [sin⁻¹(32 × 95/80²)]/2
= [sin⁻¹(3040/6400)]/2
= [sin⁻¹(0.475)]/2
= 28.36°/2
= 14.18°
Converting 14.18° to radians, we have 14.18° × π/180° = 0.2475 rad
= 0.25 rad to the nearest hundredth radian
I believe it is the final choice because it correctly compares 2 sets of corresponding sides.
Answer:
x = -6
Step-by-step explanation:
-2 = 12/x
-2x = 12
x = 12/-2
x = -6
The sum of the angles in a triangle is 180, so the equation will be :
2k+k+45= 180
add similar variables
3k+45=180
3k=135 (divide each side by 3)
k= 45
So 2k equals= 90
k=25
This is right angle triangle for your info!
Answer:
T´(z)=n^2s/4*i
Step-by-step explanation:
T(z)=d/dz(sin^2*z/4)
T´(z)=d/dz(sn^2zi/4)
T´(z)=d/dz(sn^2z/4*i)
T´(z)i*d/dz(sn^2z/4)
T´(z)=n^2s/4*i
Simplify
