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Mathematics

1 answer:

kirill115 [55]3 years ago

4 0

Answer:

the answer is 2

Step-by-step explanation:

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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

hoa [83]

$\bf \begin{array}{ccccllll}
&distance&rate(km/hr)&time(hrs)\\
&\textendash\textendash\textendash\textendash\textendash\textendash&\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash&\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\\
A&d&60&t\\
B&600-d&v&t+3
\end{array}$

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

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

8 0

3 years ago

What is the reduced and simplified value of the expression given below?

blsea [12.9K]

Answer:

$\frac{ \sqrt{16} \times \sqrt{25} } { \sqrt{4} } = \frac{4 \times 5}{2} \\ = 20 \div 2 \\ = 10$