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Tcecarenko [31]
3 years ago
11

Irina rode her bike to work at an average speed of 16 miles per hour. It started to rain , so she got a ride home along the same

route in her coworkers's car at an average speed of 27 miles per hour. If Irina's ride home in the car took 24 minutes (0.4 of an hour), how many hours was her bike ride to work, to the nearest tenth of an hot?
Mathematics
1 answer:
Lyrx [107]3 years ago
4 0
Here, the distance traveled is not given, but Irina's bike-riding speed is:  it is 16 mph.

Let the distance traveled be d.  Then d = (27 mph)(0.4 hr) = 10.8 miles.

By bicycle, distance = rate times time.  Here, 10.8 miles = (16 mph)(time), and so

                10.8 mi
time = --------------- = 0.675 hour (40.5 minutes).
             16 mph
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Answer:  7\frac{5}{7} unit and 18 unit

Step-by-step explanation:

Let ABCD is a trapezoid where AB and CD are the bases. ( In which AB is greatest base which shown in below figure)

AD and BC are the legs of the trapezoid ABCD.

Now, we have ( According to the question ),

AB = 18 unit, BC = 7 unit, AD = 3 unit and DC = 11 unit.

Here the leg AD extends from point D.

Similarly leg BC extends from point C.

Let they meet at point P ( shown in below diagram)

Since In triangles PAB and PDC,

∠PDC ≅ ∠PAB ( because DC ║ AB )

And, ∠ PAB ≅ ∠ PBA

∠DPC ≅ ∠ APB ( reflexive)

Therefore, By AAA similarity postulate,

\triangle PDC \sim \triangle PAB

Thus, By the definition of similarity,

\frac{PD}{PA} = \frac{DC}{AB}

\frac{PD}{PD+3} = \frac{11}{18} ( because PA = PD+DA)

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⇒7PD = 33

⇒ PD = 33/7

Again by the definition of similarity,

\frac{PC}{PB} = \frac{DC}{AB}

\frac{PC}{PC+7} = \frac{11}{18} ( because PB = PC + CB)

⇒ 18 PC = 11 PD +77

⇒7PC = 77

⇒ PC = 11

Thus, PA =  PD+DA = 33/7 + 3 = 7\frac{5}{7}

And, PB = PC + CB = 11 + 7 = 18


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Answer:

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