Initial velocity of the plane is Vo = 0.
acceleration a = 1.3 m/s2
total distance = 2.5 km = 2500m
time taken to reach 2.5 km with 1.3m/s^2 acceleration = t
S = Vo t + 0.5 a t^2
2500 = 0 + (0.5*1.3* t^2)
t^2 = 3846.15
t = 62 s
the maximum velocity plan can reach within 62 s is Vt
Vt = Vo + a t
Vt = 0 + (1.3*62)
Vt = 80.6 m/s
Since 80.6 m/s is greater than 75 m/s, plane can use this runway to takeoff with required speed.
Answer:
Correct choices are A and C
Step-by-step explanation:
Inscribed angles property: The inscribed angles subtended by the same arc are equal.
1. Angles EFH and EGH are both inscribed angles subtended by the arc EH. Therefore, these angles are congruent (option A is true).
2. Angles GHF and GEF are both inscribed angles subtended by the arc GF. Therefore, these angles are congruent (option C is true).
3. Angles EGH and FHG are interior angles of the triangle KGH and can be congruent (if triangle is isosceles) or can be not congruent (in general). Thus, option B is false.
4. Angles EFH and FHG in general are not congruent. They can be congruent only when arcs EH and FG have the same measure. In general, option D is false.
Isolate the variable by dividing each side by factors that don't contain the variable.
x = 1
-6+3x+4=2-4x+3
-8+7x+1=0
7x=7
x=1
Answer:
H. 95°
Step-by-step explanation:
m∠BGC is 10° smaller than m∠CGD, which means that m∠BGC is x° - 10°
m∠DGE is 25° more than m∠EGF, which means that m∠DGE is x° + 25°
The sum of all angles in a placement like this will always equal 360°
So we put all the measures of all angles on one side of thee equation and set it equal to 360*
x° + x° - 10° + 2x° + 25° + 2x° + 3x° + 30° = 360°
After combining like terms we get 9x° + 45° = 360°
After some algebra we get x = 35°
Now at the start we said that m∠DGE is x + 25°
Plug in 35° for x and we get 35° + 25°
The final answer is m∠DGE = 95°
Edit: Sorry if my answer looks a bit confusing. This is actually my first answer on Brainly so I'm quite new to this experience.
