The west constituent of their sequence needs to cancel out 58 mph crosswind. Subsequently a northwest direction is a 45-degree angle up to even with the destination. That is the third point out of the triangle and the right angle is at the destination. The top side is the west constituent of their flight the vertical side is their resultant travel and the hypotenuse is their definite distance flown. Since the 58 mph crosswind was negated by flying northwest, the distance from the beginning to the destination must be the same distance as the west component of their travel. The hypotenuse is square root of twice the side since it has 2 identical sides.
c = sqrt (58^2 + 58^2) = sqrt (6728) = 82.02
Alternative solution:
c = sqrt (2) * 58 = 1.414 * 58 = 82.02
Therefore, they have to fly 82.02 mph
Answer:
Inter Quartile Range
Explanation:
Quartile is a positional statistical average, which divided the data into 4 equal halves.
Q1 (Lower Quartile) has 25% data below it, 75% above it. Q3 (Upper Quartile) has 75% data below it, 25% above it.
Interquartile range is the measure used to calculate how far the lower & upper quartiles are.
<span>LOCATION Z, because it is only 2 away from the coast.
The rest are farther inland
hope this helps</span>
Answer:
She must be launched with minimum speed of <u>57.67 m/s</u> to clear the 520 m gap.
Step-by-step explanation:
Given:
The angle of projection of the projectile is, 
°
Range of the projectile is, 
m.
Acceleration due to gravity, 
The minimum speed to cross the gap is the initial speed of the projectile and can be determined using the formula for range of projectile.
The range of projectile is given as:
Plug in all the given values and solve for minimum speed, 
.
Therefore, she must be launched with minimum speed of 57.67 m/s to clear the 520 m gap.
