A ladder 25 feet long is leaning against the wall of a building. Initially, the foot of the ladder is 7 feet from the wall. The
foot of the ladder begins to slide at a rate of 2 ft/sec, causing the top of the ladder to slide down the wall. The location of the foot of the ladder at time t seconds is given by the parametric equations (7 +2t, 0). The location of the top of the ladder will be given by parametric equations (0, y(t)). The formula for y(t)=√[625−(7+2t)²]. The domain of t values for y(t) ranges from 0 to 9.
Calculate the average velocity of the top of the ladder on each of these time intervals (correct to three decimal places):
a) [0,2]
b) [2,4]
c) [6,8]
d) [8,9]
Calculate the average velocity of the top of the ladder on each of these time intervals (correct to three decimal places):
a) [0,2]
b) [2,4]
c) [6,8]
d) [8,9]
1 answer:
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Answer:
84°
Step-by-step explanation:
I think your question is missed of key information, allow me to add in and hope it will fit the original one. Please have a look at the attached photo.
Given that:
- ∠LOI=38°
- ∠JOP=84 °
As we can see from the figure that ∠JOP and ∠JKP are on the same segment JP. Hence, ∠JOP and ∠JKP are congruent and the angle of ∠JKP is 84°
