Distance calculation (right triangle)

AI Thread Summary
The discussion revolves around calculating the distance between city A and city D using the Pythagorean theorem. The car travels 250 miles east, 300 miles north, and 700 miles west, leading to a calculated distance of approximately 540.8 miles. There is confusion regarding the correct answer choice, with options being 500 or 600 miles. Participants suggest that the discrepancy may arise from significant figures in the measurements provided. Ultimately, the conclusion is that the professor should be consulted regarding the answer options.
future_vet
Messages
169
Reaction score
0

Homework Statement


Starting from city A, a car drives 250 miles east to city B, then 300 miles north to city C, and then 700 miles west to city D. What is the distance between city A and city D? (Points: 1)
300 mi
400 mi
500 mi
600 mi

Homework Equations


A^2 + B^2 = C^2

The Attempt at a Solution



A^2= 300^2
B^2= (700-250)^2
C^2= 292500
C= 540.8 miles.
Which answer should I choose? 500 or 600?

Thanks!
 
Physics news on Phys.org
Are you sure these are the listed answers?
 
Yes, did you get the same result I did?... Maybe it has something to do with significant figures?...
 
future_vet said:
Yes, did you get the same result I did?... Maybe it has something to do with significant figures?...

Your calculation is correct. The offered answers are obviously wrong. Mistakes happen. :wink:
 
I will ask the professor.
Thanks anyway!
 
I think that the answer is 500. He might count 300 and 700 to only have 2 significant figures, as opposed to 300. and 700. which have 3. Therefore, if the answer is 540, it would only be 500 if we take this into consideration... I think.
 

Thread 'Family of lines that are at a distance of 5 from the origin'

I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
View full post »

Thread 'Making the denominator of a certain fraction real'

Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
View full post »

Similar threads

Stuck on a question from Precalculus 2
Replies
15
Views
10K
Tintina fault (Yukon, Alaska, US and Canada) becoming more active?
Replies
2
Views
159
Trig problem involving a triangle's angles and sides
Replies
5
Views
2K
Trig Right triangle Trigonometry
Replies
2
Views
3K
How can I solve these two physics problems? (equilibrium and moment)
Replies
9
Views
1K
Related rates - finding hypotenuse of triangle
Replies
2
Views
2K
Solving for the third one should be straightforward.
Replies
10
Views
2K
Total displacement using vector diagram
Replies
3
Views
3K
Rethinking infrastructure in flood prone areas
2
Replies
56
Views
5K
Calculating Routes on a Grid: Using Combinations and Pascal's Triangle
Replies
3
Views
5K

Hot Threads

Recent Insights

Back
Top