Look into my attachment

in rhombus ABCD, ∠AOB = 90°

Look at ΔAOB
the sum of inner angles = 180°
∠BAO + ∠AOB + ∠ABO = 180°
2x + 1 + 90 + 2x + 5 = 180°
4x + 96 = 180
4x = 84
x = 21

5 0

3 years ago

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You are to drive to an interview in another town, at a distance of 270 km on an expressway. The interview is at 11:15 a.m. You p

Yuri [45]

Answer:

the least required speed is 103.88 km/h

Step-by-step explanation:

Data provided in the question:

Total distance = 270 km

Total time to reach = Time between 8 a.m and 11:15 a.m i.e 3.25 hours

Now,

For the distance of 110 km speed was 100 km/h

therefore, the time taken to cover 110 km = $\frac{\textup{Distance}}{\textup{speed}}$

= $\frac{\textup{110 km}}{\textup{100 km/h}}$

= 1.1 hour

For another 43 km speed was 42 km/h

therefore, the time taken to cover 43 km = $\frac{\textup{Distance}}{\textup{speed}}$

= $\frac{\textup{43 km}}{\textup{42 km/h}}$

= 1.0238 hours

Now,

The distance left to be covered = 270 - 110 - 43 = 117 km

Time left = 3.25 h - 1.1 h - 1.0238 h = 1.1262 h

Thus required speed = $\frac{\textup{117 km}}{\textup{1.1262 h}}$

= 103.88 km/h

Hence, the least required speed is 103.88 km/h

6 0

2 years ago