Abus takes 2 hrs less for a journey of 300 km, if its speed is increased by 5km/hr from its usual speed, then the usual speed in km/hr is

Let the usual speed of bus be x km/hr.

Time = Distance / Speed = 300/x hrs

According to the question,

=> (300/x - 2) (x + 5) = 300

=> (300 - 2x)/x (x + 5) = 300

=> 300x + 1500 - 2x² - 10x = 300x

=> 2x² + 10x - 1500 = 0

=> 2(x² + 5x - 750) = 0

=> x² + 30x - 25x - 750 = 0

=> x(x + 30) - 25(x + 30) = 0

=> (x - 25) (x + 30) = 0

=> x = 25 or x = - 30

∴ x ≠ - 30 (Because speed can't be negative)

Hence,

The usual speed of bus = 25 km/hr

9) i can't see the angle so whatever it is just do this: cos(angle) x hypotenuse and then use that answer and they hypotenuse and plug it into this: (answer you just got)^2 + b^2 = (hypotenuse)^2. the first answer will be the answer of the adjacent side and the second answer after this equation will be the opposite side.

10) y = 14 x = √147 or 12.12435565

11) x = 4.8

12) x = 4.2

13) x = 9.3

14) pq: 13.2

qr : 17.6

angle p: 53º

15) angle c: 54.3º

angle a: 35.7º

16) h = 21040

v = 1841

step-by-step explanation:

10) 7/sin(30) = 14 = y

7^2 + b^2 = 14^2

49 + b^2 = 196

b^2 = 196 - 49

√b^2 = √147

x = √147

11) cos(70) x 14 = 4.788282007

x = 4.8

12) tan(25) x 9 = 4.196768923

x = 4.2

13) sin(31) x 18 = 9.270685348

x = 9.3

14) pq: sin(37) x 22 = 13.23993051

pq: 13.2

qr: cos(37) x 22 = 17.56998122

qr : 17.6

angle p: 180 - (37 + 90) = 53º

15) cos-1(7/12) = 54.31466529

angle c: 54.3º

angle a: 180 - (90 + 54.31466529) = 35.68533471

angle a: 35.7º

16) 4 x 5280 = 21120

cos(5) x 21120

h = 21039.63202

h = 21040

sin(5) x 21120

v = 1840.729287

v = 1841

a c

step-by-step explanation: