Two bike riders left each other and started to ride in opposite directions. two hours later they were 54 miles apart. if one of them averaged twice the average rate of the other, what was the rate of each?

The speed of right going rider is 9 mph

The speed of left going rider is 18 mph

Step-by-step explanation:

Given as :

The total distance apart both the riders = 54 miles

Let The speed of right going rider = x mph

and The speed of left going rider = 2 x  mph

The Distance cover by right going rider = D miles

The Distance cover by left going rider = 54 - D  miles

Total time for both = 2 hours

So, Time =

Or ,  Distance = speed × Time

For right going rider

       D = x × 2

For left going rider

      54 - D = 2 x × 2

Or, from first equation

     54 - 2 x = 4 x

or, 54 = 4 x + 2 x

or, 6 x = 54

∴   x =

I.e x = 9 mph

So, The speed of right going rider = 9 mph

and The speed of left going rider = 2 × 9 = 18  mph

Hence The speed of right going rider is 9 mph

and The speed of left going rider is 18 mph   answer

sec ф = 5/3 (quadrant 4)

let x = base

y = perp

z = hypotenuse

sec ф = hypotenuse/base = 5/3

⇒ z = 5 and x = 3

finding y:

z² = x² + y²

5² = 3² + y²

y² = 25-9

y² = 16

y = 4

now finding all other geometric functions of theta:

sin ф = y/z = 4/5

sin ф = -4/5 (sin is negative in the 4th quadrant)

cos ф = x/z = 3/5

cos ф = 3/5 (cos is positive in 4th quadrant)

tan ф = y/x = 4/3

tan ф = -4/3 (tan is negative in 4th quadrant)

cosec ф = z/y = 5/4

cosec ф = -5/4 (cosec is negative in the 4th quadrant)

cot ф = x/y = 3/4

tan ф = -3/4 (tan is negative in 4th quadrant)