Let x be the length of the train.

On the basis of the observer;
Speed of the train = x/6

On the basis of the bridge;
Total distance covered by any point of the train= 350+x
Speed = (350+x)/20

Equating the two expressions of speed;
x/6 = (350+x)/20
20(x) = 6(350+x)
20x = 2100+6x
(20-6)x = 2100
14x = 2100
x= 2100/14 = 150 m
Speed = x/6 = (350+x)/20 = 150/6 = 500/20 = 25 m/s.

Therefore,
Length of train = 150 m
Speed of train = 25 m/s

8 0

3 years ago

What is the volume of the composite figure? Round your answer to the nearest hundredth. Use 3.14 for π.

Reil [10]

Answer:

263.76 in^3

VOLUME OF FIGURE=VOL.OF CONE+VOL.OF CYLINDER +VOL.OF HEMISPHERE

=(1/3×3.14×3×3×4)+(3.14×3×3×6)+(2/3×3.14×3×3×3)

=37.68+169.56+56.52

=263.76in^3

7 0

3 years ago

What is the product?

fredd [130]

Answer:

$\frac{4k + 2}{k^{2}-4 }$ × $\frac{k-2}{2k+1}$ = $\frac{2}{k + 2}$

Step-by-step explanation:

$\frac{4k + 2}{k^{2}-4 }$ × $\frac{k-2}{2k+1}$

To solve the above, we need to follow the steps below;

4k+2 can be factorize, so that;

4k +2 = 2 (2k + 1)

k² - 4 can also be be expanded, so that;

k² - 4 = (k-2)(k+2)

Lets replace 4k +2 by 2 (2k + 1)

and

k² - 4 by (k-2)(k+2) in the expression given

$\frac{4k + 2}{k^{2}-4 }$ × $\frac{k-2}{2k+1}$

$\frac{2(2k+ 1)}{(k-2)(k+2)}$ × $\frac{k-2}{2k+1}$

(2k+1) at the numerator will cancel-out (2k+1) at the denominator, also (k-2) at the numerator will cancel-out (k-2) at the denominator,

So our expression becomes;

$\frac{2}{k + 2}$

Therefore, $\frac{4k + 2}{k^{2}-4 }$ × $\frac{k-2}{2k+1}$ = $\frac{2}{k + 2}$

3 0

3 years ago

Read 2 more answers

A trapeziodal pyramid has a volume 2856insquared the height of the pyramid is 24in and the lengths of the 2 two bases of the tra

zalisa [80]

V = 1/3Bh
B = 1/2(b1 + b2)t

V = 1/3[1/2(b1 + b2)t] * h
h = 24
b1 = 13
b2 = 29
t = ?
V = 2856

Substitute:
2856 = 1/3[1/2(13 + 29)t](24)
2856 = 1/6(42)(24)t
2856 = 7(24)t
2856 = 168t
t = 2856/168 = 17 in height of the trapeziod

8 0

3 years ago