For the last part, you have to find where
attains its maximum over
. We have

so that

with critical points at
such that





So either

or

where
is any integer. We get 8 solutions over the given interval with
from the first set of solutions,
from the set of solutions where
, and
from the set of solutions where
. They are approximately






I had to read this a couple of times to see this was not about time, but is a right triangle problem. If the hour hand is at three, let us consider that a leg of a right triangle, we will consider the hypotenuse as the second hand.
Sketch the picture.
The hypotenuse is 9 cm. the angle is found by:
the (hour hand) is pointing directly at the 12. I expect you mean the minute hand. The angle between the minute hand and the second hand is 25/60*360 = 150 which make the angle between the hour hand and the second hand. 150 -90 =60
so the second hand and the hour hand gives us a right triangle, with a 60 degree angle and a hyp. of 9 cm.
cos 60 = x/9
9 cos 60 =x
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x = 4.5 cm
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0.17 sense you have to round it to the nearest hundreth
Answer:
x = 25/4
Step-by-step explanation:
Because of the known similarity of the triangles, we know that
10 x
----- = ----
16 10
Cross-multiplying, we get 16x = 100, and thus x = 100/16 = 50/8 = 25/4
x = 25/4