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Ipatiy [6.2K]
3 years ago
15

You are traveling 180 miles back to your home town for a class reunion. About 60 miles of the trip are through areas where the s

peed limit is 45 miles per hour and the rest of the trip is through areas where the speed limit is 55 miles per hour . Assuming that you can travel at the speed limits to get the reunion, how long will it take you? Round your answer to the nearest tenth
Mathematics
1 answer:
kicyunya [14]3 years ago
7 0

Answer:

2.5hour

Step-by-step explanation:

by adding both and substracting from total miles

Something
2 years ago
can you show the work necessary
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Ship A receives a distress signal from the north, And ship B receives a distress signal from the same vessel from the southeast.
11Alexandr11 [23.1K]

The coordinates of Ship A and B are missing, so i have attached it

Answer:

The vessel in distress is located at the coordinate (1, 2)

Step-by-step explanation:

The coordinates attached shows that;

Coordinates of A are (3, 4) while that of B are (1, 1).

We are told that Ship A receives a distress signal from the north, And ship B receives a distress signal from the same vessel from the southeast.

Now, from the image attached, If we imagine extending the line of ship A that is getting distress signal from southeast and also same thing for ship B that is getting signal from north,we'll discover that the lines intersect at a point with co-ordinates of approximately; (1, 2)

8 0
2 years ago
What is the sum of the first 70 consecutive odd numbers? Explain.
expeople1 [14]

The sum of the first n odd numbers is n squared! So, the short answer is that the sum of the first 70 odd numbers is 70 squared, i.e. 4900.

Allow me to prove the result: odd numbers come in the form 2n-1, because 2n is always even, and the number immediately before an even number is always odd.

So, if we sum the first N odd numbers, we have

\displaystyle \sum_{i=1}^N 2i-1 = 2\sum_{i=1}^N i - \sum_{i=1}^N 1

The first sum is the sum of all integers from 1 to N, which is N(N+1)/2. We want twice this sum, so we have

\displaystyle 2\sum_{i=1}^N i = 2\cdot\dfrac{N(N+1)}{2}=N(N+1)

The second sum is simply the sum of N ones:

\underbrace{1+1+1\ldots+1}_{N\text{ times}}=N

So, the final result is

\displaystyle \sum_{i=1}^N 2i-1 = 2\sum_{i=1}^N i - \sum_{i=1}^N 1 = N(N+1)-N = N^2+N-N = N^2

which ends the proof.

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Step-by-step explanation:

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