Answer:
one million two hundred thirty four thousand six hundred seventy eight
<u>Given</u>:
Given that the isosceles trapezoid JKLM.
The measure of ∠K is 118°
We need to determine the measure of each angle.
<u>Measure of ∠L:</u>
By the property of isosceles trapezoid, we have;



Thus, the measure of ∠L is 62°
<u>Measure of ∠M:</u>
By the property of isosceles trapezoid, we have;

Substituting the value, we get;

Thus, the measure of ∠M is 62°
<u>Measure of ∠J:</u>
By the property of isosceles trapezoid, we have;

Substituting the value, we get;

Thus, the measure of ∠J is 118°
Hence, the measures of each angles of the isosceles trapezoid are ∠K = 118°, ∠L = 62°, ∠M = 62° and ∠J = 118°
The labeled point in which it will be rotated around
Normally you would only solve by completing the square if a is 1 and b is even. But the answer would be half of b. So half of 9 is 4.5 or 4(1/2).
Answer:
The cosine function to model the height of a water particle above and below the mean water line is h = 2·cos((π/30)·t)
Step-by-step explanation:
The cosine function equation is given as follows h = d + a·cos(b(x - c))
Where:
= Amplitude
2·π/b = The period
c = The phase shift
d = The vertical shift
h = Height of the function
x = The time duration of motion of the wave, t
The given data are;
The amplitude
= 2 feet
Time for the wave to pass the dock
The number of times the wave passes a point in each cycle = 2 times
Therefore;
The time for each complete cycle = 2 × 30 seconds = 60 seconds
The time for each complete cycle = Period = 2·π/b = 60
b = π/30 =
Taking the phase shift as zero, (moving wave) and the vertical shift as zero (movement about the mean water line), we have
h = 0 + 2·cos(π/30(t - 0)) = 2·cos((π/30)·t)
The cosine function is h = 2·cos((π/30)·t).