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Zielflug [23.3K]
3 years ago
9

I REALLY NEED YOUR HELP ANSWER THIS QUESTION IN FULL OR ILL BE VERY UPSET :/

Mathematics
1 answer:
dusya [7]3 years ago
8 0
X>or=0
√(150x^3)...........
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How do you solve (Arc)QPT if <QZT = 120
yuradex [85]
By definition, the arc length is given by:
 arc = R * theta * ((2 * pi) / 360)
 Where,
 theta: angle in degrees
 R: radio
 We have then:
 (Arc) QPT if <QZT = 120:
 theta = 360-120 = 240 degrees
 R = 13.5 units
 Substituting values we have:
 (Arc) QPT = R * theta * ((2 * pi) / 360)
 (Arc) QPT = (13.5) * (240) * ((2 * pi) / 360)
 (Arc) QPT = 56.55 units
 Answer:
 
(Arc) QPT = 56.55 units
7 0
4 years ago
-14n + q=rt-4n, for n
pentagon [3]
N = -1/10rt + 1/10q
= −14n+q+4n=rt−4n+4n
= −10n+q=rt
= −10n+q+−q=rt+−q
= −10n=rt−q
= -10n/-10 = rt-q/-10
So the answer is:
= n = -1/10rt + 1/10q
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