Use the three steps to solve the problem.

two planes leave an airport at noon. if the eastbound plane flies at 660 mph and the westbound plane flies at 600 mph, at what time will they be 2,000 miles apart? round your answer to the nearest minute.

{
a0 p. m.}

At 1:53 pm They will be 2,000 miles apart

Step-by-step explanation:

Given :The eastbound plane flies at 660 mph and the westbound plane flies at 600 mph

To Find : At what time will they be 2,000 miles apart?

Solution :

The eastbound plane flies at 660 mph

Suppose after time t they will be  2,000 miles apart

So, distance traveled at this time by eastbound = Distance = Speed * time

Distance = 660t

The westbound plane flies at 600 mph

Suppose after time t they will be  2,000 miles apart

So, distance traveled at this time by west bound = Distance = Speed * time

Distance = 600t

So, to calculate t time at which they will be 2000 miles apart

Since 1 hr = 60 minutes

So, 0.58 hours = 0.58*60 = 34.8 minutes

Thus after 1 hour 34 minutes they will be 2000 miles apart

Since they start at 12 noon

So, after 1 hour 53 minutes time will be 1:53 pm

Hence At 1:53 pm They will be 2,000 miles apart

the answer is (4,3)

since the rule for r0, 90º is (-y,x) and it was already applied to form the image, all you have to do is reverse the answer using this formula.

5x + 3y = - 4

step-by-step explanation:

the equation of a line in standard form is

ax + by = c ( a is a positive integer and b, c are integers )

express the line first of all in slope- intercept form

y = mx + c ( m is the slope and c the y-intercept )

to calculate m use the gradient formula

m = ( y₂ - y₁ ) / ( x₂ - x₁ )

with (x₁, y₁ ) = (7, - 3) and (x₂, y₂ ) = (4, - 8)

m = = -

y = - x + c ← is the partial equation

to find c substitute either of the 2 points into the partial equation

using (4, - 8 ), then

- 8 = - + c ⇒ c = -

y = - x - ← in slope-intercept form

multiply all terms by 3 to eliminate the fractions

3y = - 5x - 4 ( add 5x to both sides )

5x + 3y = - 4 ← in standard form