Answer:
- Distance in 47 minutes = 0.75 × 47 = 35.25 miles
- Distance in 1 hour or (60 minutes) = 0.75 × 60 = 45 miles
Step-by-step explanation:
Given that the car is traveling at a steady speed.
Distance traveled in 2 1/3 or 7/3 or (2.33) minutes = 1 3/4 miles
= 7/4 or 1.75 miles
Thus,
distance in 1 minute: 1.75/2.33 = 0.75 miles per minute
Thus,
- Distance in 47 minutes = 0.75 × 47 = 35.25 miles
- Distance in 1 hour or (60 minutes) = 0.75 × 60 = 45 miles
Answer:
976
Step-by-step explanation:
would you stop it already
A. As sun peeks, it creates a straight line- sun to tree to shadow, angle=180°
b. During morning, the sun-tree-shadow changes from 180° to 90°, during noon, angle =90°.During sunset the angle changes from 90 to 0°.
c. Acute angles created in afternoon.
d. A right angle would be created at noon.
e. Obtuse angle would be created in morning.
f. Yes, a straight angle would be created at sunrise.
g. When sun is at mid-morning, the angle=45°+90°= 135°
Answer:
W = ∫ [−2/3 to 0] 66π (2−y)(1−y²) dy = 2332π/27
Step-by-step explanation:
Vertical cross-section of bowl is a semi-circle of radius 1, centered at (0,0)
x² + y² = 1 (for y ≤ 0)
Bottom of bowl is at y = −1
Top of bowl is at y = 0
When punch is 4 inches deep, this 1/3 ft above bottom of bowl:
y = −1+1/3 = −2/3
So we integrate from y = −2/3 to y = 0
Radius of cross section at height of y is = x = √(1−y²)
Area of cross section at height y is πr²= πx² = π(1−y²)
Since each person is to lift ladle 2 feet above the top of the bowl, then distance = 2−y
W = Σ (density of punch * distance * volume of punch)
W = ∫ [−2/3 to 0] 66 * (2−y) * πr² dy
W = ∫ [−2/3 to 0] 66π (2−y)(1−y²) dy = 2332π/27
Answer:
1/100
Step-by-step explanation:
The answer is 1/100 because we don't know how many people are in the group but do know that only 1 person has been taken from the group and the probability will be 1 out of how many people are in the group so we just put 100 there as we don't know how many people are in the group.
<em>Ps: Sorry if this is wrong, this is just a guess.</em>