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VikaD [51]
3 years ago
14

 The distance between city A and B is 600 km. The first train left A and headed towards B at the speed of 60 km/hour. The secon

d train left B heading towards A three hours after the first train left A, and it traveled with a speed of v km/hour. The trains met t hours after the time at which the first train left A. Express v in terms of t. Find the speed v if t=7; t=6.
Mathematics
1 answer:
hoa [83]3 years ago
8 0
\bf \begin{array}{ccccllll}
&distance&rate(km/hr)&time(hrs)\\
&\textendash\textendash\textendash\textendash\textendash\textendash&\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash&\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\\
A&d&60&t\\
B&600-d&v&t+3
\end{array}

\bf \textit{meaning}\implies 
\begin{cases}
d=(60)(t)
\\ \quad \\
600-d=(v)(t+3)\\
------------\\
d=\boxed{60t}\qquad thus
\\ \quad \\
600-\boxed{60t}=v(t+3)\leftarrow \textit{solve for "t"}
\end{cases}

keep in mind, that "t" is the time when the train at A station, left towards B station

they met, at some time "t", and by the time that happened, train from A
which started 3 hours earlier, had already covered "d" distance,
whatever that is
and the train coming from B, covered, 600-d, or the difference
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7 0
3 years ago
Read 2 more answers
An arched drainage culvert can be modelled by the function:
Zanzabum

Substitute 136.5 for y.

136.5=\frac{x(191-x)}{60}

8190=x(191-x)

8190=191x-x^{2}

x^{2} -191x+8190=0

Compare this equation with ax^{2} +bx+c=0

a = 1, b = - 191, c = 8190

x=\frac{-b+\sqrt{b^{2}-4ac } }{2} or

x=\frac{-b-\sqrt{b^{2}-4ac } }{2}

x=\frac{191+\sqrt{191^{2}-4(1)(8190) } }{2} or

x=\frac{191-\sqrt{191^{2}-4(1)(8190) } }{2}

x=\frac{191+\sqrt{36481-32760 } }{2} or

x=\frac{191-\sqrt{36481-32760 } }{2}

x=\frac{191+\sqrt{3721 } }{2} or

x=\frac{191-\sqrt{3721 } }{2}

x=\frac{191+61}{2} or

x=\frac{191-61}{2}

x=\frac{252}{2} or

x=\frac{130}{2}

x = 126 or x = 65

Hence, the points on the curve are (65, 136.5) and (126, 136.5).

The width of the air space is the distance between these points.

Width = \sqrt{(x_{2}- x_{1})^{2} +( y_{2}- y_{1}) ^{2}

= \sqrt{(126- 65)^{2} +(136.5-136.5) ^{2}

= 126 - 65

= 61

Hence, width of the air space is 61m.



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3 years ago
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