How many miles from ft Lauderdale to Haiti

Answers

Answer 1
Answer:
Ft. Lauderdale is a city that stretches several miles, and Haiti
is a whole country, that's many miles wide and many miles high. 

In order to nail down a reliable answer, you'd really need to specify
one point in Ft. Lauderdale and one point in Haiti.

If you start at the northwest end of Runway-13 at Ft. Lauderdale
Executive Airport, and take the shortest possible route to the east
end of Runway-28 at the Aeroport International de Port au Prince
at Haiti's capital city, you'd have to travel 727.57 miles.

But if you start in Ft. Lauderdale at the intersection of Griffin Rd
and Ravenswood Rd, and take the shortest possible route to the
Dispensaire de Bord-de-Mer hospital on Haiti's north coast, you'd
only have to travel 613.63 miles.

You really need to say WHERE in Ft. Lauderdale and WHERE in Haiti.
Answer 2
Answer: ♥ Based upon research I found that 
♥ From Ft. Lauderale to Haiti is exactly 703 miles 

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Which type of triangle is formed with the points A(1, 7), B(-2, 2), and C(4, 2) as its vertices?

Which of the following figures are quadrilaterals?

Answers

square , rectangle , rombus , parallelogram

Answer:

a four-sided figure.

Step-by-step explanation:

What is -4x=-2+4y -4y=1-4x

Answers

I hope this helps you

Zeb and his wife have a taxable income of $167,487. What is their tax liability?A. $46,896.36
B. $73,583.86
C. $8,452.36
D. $35,139.86

Answers

Answer:

D

Step-by-step explanation:

What is the period and frequency of a 10-pound weight oscillating (as a pendulum) on a 5-meter-long rope?

Answers

The mass of the pendulum does not affect the period and frequency
The period can be determined using the formula:
T = 2π √(L/g)

where L is the length of the rope
and g is the acceleration due to gravity = 9.8 m/s2

Plugging in the values:
T = 2π √(5/9.8)
T = 10π/7 = 4.49 s

And for the frequency:
f = 1/T
f = 1/4.49
f = 0.22 Hz

Answer:

Period = 4.5 seconds

Frequency = 0.22 hertz.

Step-by-step explanation:

Given : 10-pound weight oscillating (as a pendulum) on a 5-meter-long rope.

To find : What is the period and frequency.

Solution :

The formula to find time when length is given,

T=2\pi\sqrt{(L)/(g)}

where,T is the time period , L is the length and g is the acceleration due to gravity

L= 5 m and g=9.8 m/s

Substitute in the formula,

T=2\pi\sqrt{(5)/(9.8)}

T=2\pi√(0.510)

T=2* 3.14* 0.714

T=4.483

The time period is approx 4.5 seconds.

\text{Frequency}=\frac{1}{\text{Period}}

\text{Frequency}=(1)/(4.5)

\text{Frequency}=0.22\text{hertz}

Therefore,Period = 4.5 seconds

Frequency is 0.22 hertz.

(37x+9)+(32x+2) whats the answer

Answers

Answer:

69x+11

Step-by-step explanation:

What is the equation of the line, in slope-intercept form, that contains the points (−1, 5) and (2, 2)?y = −3x + 4
y = −x + 1
y = x + 4
y = −x + 4

Answers

So,

First, we need to find the slope by subtracting the y-values and subtracting the x-values.

5 - 2 = 3
-1 - 2 = -3

Now divide the y-result by the x-result.
3/-3 = -1 <-- slope

y = -x + b <-- equation so far

To find the y-intercept, substitute a solution into the equation.

2 = -2 + b

Add 2 to both sides.
4 = b

Now we have our complete equation.
y = -x + 4 <-- option D
y=mx+b
m=slope
b=yintercept
slope=(y2-y1)/(x2-x1)

given the points (-1,5) and (2,2)
slope=(2-5)/(2-(-1))=-3/(2+1)=-3/3=-1
y=-1x+b
sub a oint
(2,2)
x=2
y=2
2=-1(2)+b
2=-2+b
add 2
4=b

y=-x+4