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3 years ago

6

Find the average speed of a plane that leaves Bristol at 14.00 and lands in Rome at 17.15.The distance is 1200 miles

1 answer:

horrorfan [7]3 years ago

6 0

It is very important to go through all the details that have been provided in the question. These information's will come in very handy while getting to the solution for the answer. Now let us concentrate on the problem in hand.
Distance between Bristol and Rome that the plane covers = 1200 miles
Starting time of the plane from Bristol = 14:00 hrs
Landing time of the plane in Rome = 17:15 hrs
Then
Total time taken by the plane to travel from Bristol to Rome = 17:15 - 14:00 hours
= 3 .15 hours
So
The average speed of the plane = 1200/3.15 miles per hour
= 380.95 miles per hour
From the above deduction we can conclude that the average speed of the plane traveling from Bristol to Rome is 380.95 miles per hour.

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Answer:

Step-by-step explanation:

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3 years ago

A bridge in the shape of an arch connects two cities separated by a river. The two ends of the bridge are located at (–7, –13) a

sdas [7]

Answer:

$y=-\dfrac{13}{49}x^2$

Step-by-step explanation:

The shape of an arch corresponds to a parabola.

the general equation for a parabola is:

$y=ax^2+bx+c$

we're given three coordinates: (-7,-13),(7,-13) and (0,0)

so we can plug these values in the general equation to make 3 separate equations:

(x,y) = (-7,-13)

$-13=a(-7)^2+b(-7)+c$

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(x,y) = (7,-13)

$-13=a(7)^2+b(7)+c$

$49a+7b+c=-13$

(x,y) = (0,0)

$0=a(0)^2+b(7)+c$

$c=0$

so we have three equations. and we can solve them simultaneously to find the values of a,b, and c.

we've already found c = 0, let's use substitute it to other equations.

$49a-7b+c=-13\quad\Rightarrow\quad49a-7b=-13$

$49a+7b+c=-13\quad\Rightarrow\quad49a+7b=-13$

we can solve these two equation using the elimination method, by simply adding the two equations

$\quad\quad49a-7b=-13\\+\quad49a+7b=-13$

------------------------------

$\quad\quad 98a=-26$

$\quad\quad a=-\dfrac{13}{49}$

Now we can plug this value of a in any of the two equations.

$49a-7b=-13$

$49\left(-\dfrac{13}{49}\right)-7b=-13$

$-13-7b=-13$

$-7b=0$

$b=0$

We have the values of a,b, and c. We can plug them in the general equation to find the equation of the arch.

$y=\left(-\dfrac{13}{49}\right)x^2+0x+0$

$y=-\dfrac{13}{49}x^2$

$49y=-13x^2$

This our equation of the arch!

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3 years ago

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Aleksandr-060686 [28]

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Answer:

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Given data

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3 years ago