The projectile's horizontal and vertical positions at time 
are given by
where 
. Solve 
for the time it takes for the projectile to reach the ground:
In this time, the projectile will have traveled horizontally a distance of
The projectile's horizontal and vertical velocities are given by
At the time the projectile hits the ground, its velocity vector has horizontal component approx. 176.77 m/s and vertical component approx. -178.43 m/s, which corresponds to a speed of about 
.
The answer is 0.50.
P(4<=X<=8)=P(x=4)+ P(x=5)+ P(x=6)+ P(x=7)+ P(x=8)= 0.05+ 0.15+0.15+0.15 +0 = 0.50
Notice that in 8 the line touches the x axis so it’s corresponding probability is 0.
Answer:
35/8 if its simple multiple then
Step-by-step explanation:
7/4*5/2
35/8
Answer:
a solution is 1/2 *tan⁻¹ (2*y) = - tan⁻¹ (x²) + π/4
Step-by-step explanation:
for the equation
(1 + x⁴) dy + x*(1 + 4y²) dx = 0
(1 + x⁴) dy = - x*(1 + 4y²) dx
[1/(1 + 4y²)] dy = [-x/(1 + x⁴)] dx
∫[1/(1 + 4y²)] dy = ∫[-x/(1 + x⁴)] dx
now to solve each integral
I₁= ∫[1/(1 + 4y²)] dy = 1/2 *tan⁻¹ (2*y) + C₁
I₂= ∫[-x/(1 + x⁴)] dx
for u= x² → du=x*dx
I₂= ∫[-x/(1 + x⁴)] dx = -∫[1/(1 + u² )] du = - tan⁻¹ (u) +C₂ = - tan⁻¹ (x²) +C₂
then
1/2 *tan⁻¹ (2*y) = - tan⁻¹ (x²) +C
for y(x=1) = 0
1/2 *tan⁻¹ (2*0) = - tan⁻¹ (1²) +C
since tan⁻¹ (1²) for π/4+ π*N and tan⁻¹ (0) for π*N , we will choose for simplicity N=0 . hen an explicit solution would be
1/2 * 0 = - π/4 + C
C= π/4
therefore
1/2 *tan⁻¹ (2*y) = - tan⁻¹ (x²) + π/4
The four quadrants and the sign of coordinates is given as follows:
Quadrant I : (n, n ) both x and y positive.
Quadrant II : ( -n,n) = x negative and y positive.
Quadrant III : ( -n,-n) = both x and y negative.
Quadrant IV: (n, -n) = x positive and y negative.
So Based on the above concept , we can say that vertex C (-n,-n) has both x and y negative and so it lies in quadrant III.
Answer is Vertex C (-n,-n)
