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
Answer:
The second one
Step-by-step explanation:
Step-by-step explanation:
because because that would be subtracting positive 10 by -10. not adding -10 + -10 which would be 20
Answer:
The point is in the 4th quadrant.
Explanation:
Given the ordered pair(4, -5), we have that x = 4 and y = -5. Plotting this point, we'll have;
Note that quadrants are labelled in an anti-clockwise direction with the top right portion of the graph as the 1st quadrant. Looking at the plotted point, we can see that it i in the 4th quadrant.
1. -3x + -6
2. -3x + 9
3. 2x - 6
4. -2x + 6
here u go
