Stewart Calculus, 7th Edition
Ch. 12
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Published: 2012
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Robert Jacobson
4.8
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Work
Explanation
Explanation

Please see your textbook for the exercise prompt.

Result
5 Comments
Dalilmath (/profile/Dalilmath/)
2 yrs
Sure this took a long time, good job on explaining everything clearly and not leaving any steps
out.
mohamid92 (/profile/mohamid92/)
2 yrs
we will not have it in exam ... Hopefully !!
flying_oliver (/profile/flying_oliver/)
1 yr
x2^^
Jack Trades (/profile/jackoftrades/)
1 yr
Great work! Thank you for taking the time to explain all of this!
Allen Fancydog (/profile/Allenexige/)
11 mths
Shouldn't angle PRQ = 57 degrees?
48+75+58= 181 otherwise great answer, helped me alot!
Edit: I used a cheat where I did 180­ QPR and RQP which gave me a rounded off number, i see
why you got 58 now
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eugen289 (/profile/eugen289/)
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1.6
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Explanation
Work
Explanation
Please see your textbook for the exercise prompt.
764a7d2beee20a39e167e11f/untitled.bmp)
Result
6 Comments
blorangepurplink
2 yrs
I do not understand. Why are you doing things this way? where does the 4/1 come from? Why
do some have pi and some have pi/2?
SnoopyFish (/profile/SnoopyFish/)
2 yrs
The reason the answer is like this is because adding of angles. Angle p plus the angles on
either side of it add up to 180°. Pi = 180°. So, 180° ­ the angle left of P ­ the angle to the right of
P = P.
fuzzylogic11 (/profile/fuzzylogic11/)
2 yrs
The arctg 4/1 for angle Q should be arctg 3/1
Alex Cares
2 yrs
I agree with fuzzy
vermindo (/profile/vermindo/)
2 yrs
This is why there is an EXPLANATION column!
markrodriguez0 (/profile/markrodriguez0/)
2 yrs
The way i solved this one was finding the lengths of pq qr and pr then split the triangle in half to
convert it into two right triangles using pathagorean theorem. Then I used trig to get the angels
for p and r and found q by subtracting 180 from p r
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