Answer:
480 km
Step-by-step explanation:
Answer:
The change in the angle of elevation is approximately 63.435°
Step-by-step explanation:
The given parameters are;
The speed of the tightrope walker = 3 ft./sec
The altitude at which the tightrope walker is moving = 100 ft.
The horizontal distance of the tightrope walker from the searchlight = 200 feet
The final angle of elevation, θ, is given by the trigonometric ratio as follows;
Whereby the tightrope walker was initially vertically overhead the search light, the initial angle of elevation = 90°
The change in the angle of elevation = The initial angle of elevation - The final angle of elevation
Substituting the values gives
The change in the angle of elevation = 90° - 26.565 ≈ 63.435°
The original equation:
Integral of csc(5x)
Use a u sub:
u = 5x
du = 5dx
Simplify the du:
Apply to the equation:
Integral csc(u)
du
Simplify:
Integral
du
Factor out the constant:
Integral
du
Use a second Substitution:
v = tan(
)
du =
dv
Applying to the equation:
Simplify:
Integrate:
Insert back in your v and u:
v = tan(
)
u = 5x
This gives us the final equation (don't forget your constant):
(Thank you for making me write it out, I made a mistake on the original answer.)
To find the z-score for a weight of 196 oz., use
A table for the cumulative distribution function for the normal distribution (see picture) gives the area 0.9772 BELOW the z-score z = 2. Carl is wondering about the percentage of boxes with weights ABOVE z = 2. The total area under the normal curve is 1, so subtract .9772 from 1.0000.
1.0000 - .9772 = 0.0228, so about 2.3% of the boxes will weigh more than 196 oz.
Answer:
Step-by-step explanation:
The angle forms arc . The measure of this arc is twice the measure of angle , since is a point on the circle. Since forms an arc , we can set up the following equation:
.
Since arc TU represents half of a circle and there are 360 degrees in a circle, we can substitute the following values and solve for the measure of arc VU: