Use the trig identity
2*sin(A)*cos(A) = sin(2*A)
to get
sin(A)*cos(A) = (1/2)*sin(2*A)
So to find the max of sin(A)*cos(A), we can find the max of (1/2)*sin(2*A)
It turns out that sin(x) maxes out at 1 where x can be any expression you want. In this case, x = 2*A.
So (1/2)*sin(2*A) maxes out at (1/2)*1 = 1/2 = 0.5
The greatest value of sin(A)*cos(A) is 1/2 = 0.5
Answer:
m(∠C) = 18°
Step-by-step explanation:
From the picture attached,
m(arc BD) = 20°
m(arc DE) = 104°
Measure of the angle between secant and the tangent drawn from a point outside the circle is half the difference of the measures of intercepted arcs.
m(∠C) = ![\frac{1}{2}[\text{arc(EA)}-\text{arc(BD)}]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5B%5Ctext%7Barc%28EA%29%7D-%5Ctext%7Barc%28BD%29%7D%5D)
Since, AB is a diameter,
m(arc BD) + m(arc DE) + m(arc EA) = 180°
20° + 104° + m(arc EA) = 180°
124° + m(arc EA) = 180°
m(arc EA) = 56°
Therefore, m(∠C) = 
m(∠C) = 18°
d = distance between the two cities
v₁ = average speed while going from chicago to kansas city = 440 knots
t₁ = time taken to travel distance going from chicago to kansas city
time taken to travel distance going from chicago to kansas city is given as
t₁ = d/v₁
t₁ = d/440 eq-1
v₂ = average speed while going from kansas city to chicago = 110 knots
t₂ = time taken to travel distance going from kansas city to chicago
time taken to travel distance going from kansas city to chicago is given as
t₂ = d/v₂
t₂ = d/110 eq-2
Given that :
t₂ = t₁ + 3
using eq-1 and eq-2
(d/110) = (d/440) + 3
d = 440
For the second one 1 mph equals 1.6 km so i’m pretty sure it would be all of them
Answer:
there's nothing here to answer
