Answer:
Plane B is closer.
The distance for Plane A is 7.9 km.
The distance for Plane B is 7.4 km.
Step-by-step explanation:
To find the distance to the tower, altitude must be converted to kilometers. Each 1000 ft is 0.3048 km, so the heights of the planes are ...
-- Plane A: (20 thousand ft)*(0.3048 km/thousand ft) = 6.096 km
-- Plane B: (8 thousand ft)*(0.3048 km/thousand ft) = 2.4384 km
The Pythagorean theorem can be used to find the distance from each plane to the tower:
-- distance = √((ground distance)² + (height)²)
-- Plane A distance = √(5² +6.096²) ≈ √62.16 ≈ 7.9 . . . km
-- Plane B distance = √(7² +2.4384²) ≈ √54.95 ≈ 7.4 . . . km
The distance for Plane B is shorter, so Plane B is closer to the tower.
Answer:
D.
Step-by-step explanation:
Answer:
B, 2(10^4)
Step-by-step explanation:
2(10^3)= 2000
2(10^4)= 20000
2(10^4) Is closer to 16790 than 2(10^3)
Answer:
Step-by-step explanation:
1). m∠AEC = m∠AEB + m∠BEC
= 21° + 37°
= 58°
2). m∠BED = m∠BEC + m∠CED
= 37° + 44°
= 81°
3). m∠IKF = m∠IKH + m∠HKG + m∠GKF
= m∠IKH + m∠HKG + m∠IKH [Since, ∠IKH ≅ ∠GKF]
= 2∠IKH + m∠HKG
103° = 2∠IKH + 41°
2(∠IKH) = 103 - 41
m(∠IKH) = 31°
4). m∠AED = m∠AEB + m∠BEC + m∠CED
= 21° + 37° + 44°
= 102°
5). m∠JKG = 108°
m∠JKG = m∠JKI + m∠IKH + m∠HKG
108° = m∠JKI + 31° + 41°
m∠JKI = 108° - 72°
m∠JKI = 36°
6). m∠HKF = m∠GKF + m∠HKG
= m∠IKH + m∠HKG [Since, m∠GKF = m∠IKH]
= 31° + 41°
= 72°
7). m∠NQO = m∠MQN = 64°
8). m∠JKF = m∠JKI + m∠IKF
= 36° + 103°
= 139°
8). m∠MQO = 2(m∠NQO)
= 2(64)°
= 128°
9). m∠LQO = 156°
m∠LQM = m∠LQO - m∠MQO
= 156° - 128°
= 28°
10. m∠NQP = m∠NQO + m∠OQP
= 64° + m∠LQM [Since ∠OQP ≅ ∠LQM]
= 64° + 28°
= 92°
Answer:
that x is represented by 3 as the value
Step-by-step explanation:
