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A 1 km long train travelling at a speed of 60km/h enter a tunnel 1km long. What time will the train take to come out of the tunnel fully.​

Hello Guest!

$t=\dfrac{s}{v}=\dfrac{(1km+1km)\cdot h}{60km}\cdot \dfrac{60min}{h}=\color{blue}2min$

The train passed the tunnel after 2 minutes.

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Standard formula for a circle

(x-h)²  + (y-k)²   = r²    

SO for this equation   r² =12      so r = sqrt(12) = 2 sqrt 3

distance / time = average speed

distance = rate * time

     time = distance/rate

(2 miles + 1 mile)    /   (2 miles/3 m/hr   + 1 mile/ 8 m/hr)

3  / (2/3 + 1/8)  = 3   /    (16+3)/24   = 3 * 24/19 = 3.79 m/hr avg

b) no  =====>  ( 3 + 8)/2 = 5.5 m/hr   does not equal   3.79 m/hr

The radius of the circle is 2.

There are 2 solutions.

The number of possible matchups is 144.

= 12000 (1 + .035)⁷  = ~ 15 267

The sum of all possible values of x is 13.

Add the individual shapes' areas

square    16 x 16

triangle = 1/2 b * h = 1/2 * 16 * 7    (I think it is 7....the pic is cut off)

semicircle = 1/2 pi r^2 = 1/2 pi (8^2)

great job guest

The distance between the two points given is the radius of the circle

(0-2)^2 + (-8- - 5)^2 = d^2     d = sqrt(13)    (this is the radius of the circle)

circle equation     (x-0)^2 + (y+8)^2 = sqrt(13)^2

                               x^2 + (y+8)^2 = 13

Δ ABS is an isosceles triangle.

∠B = 2 [arctan(3 / 4)] = 73.73979529º

AD = tanB * BD = 13.71428571

Area = AD * BD = 54.85714286 u2

Find hypot via pythag theorem

2^2 + sqrt (19)^2 = hypot^2     Hypot = sqrt(23)

Sin L = opp/hypot  = 2 /sqrt(23)

The minimum distance is 8 feet.

It's just on web2.0.

I think the answer is 28

(x+10)^2   + (y-7)^2 = 65

center h,k   is    -10,7

There are 8 integers in the sequence, as 25 ÷ 3 has a decimal and integers are whole numbers

h,k    =  1,-4

(x-1)^2 + (y+4)^2 = 3^2

2 x 14.5 khz = 29 khz sampling rate

2048 can be represented as 2¹¹         so 11 bits will be able to encode 2048 levels

29000 samples / sec  *  11 bits/sample = 319000 bits/sec  = 319 kb/s