A plane is descending into an airport. When its altitude is 3 1/2 miles, it is 250 miles from the airport. When its altitude is 1 mile, it is 75 miles from the airport. At 2 1/2 miles, it is 175 miles from the airport. Problem 6 a. If the pilot follows the same pattern, what will the plane's altitude be at 25 miles from the airport? *

1/2 miles

Step-by-step explanation:

Given the following:

Altitude (miles) - - - - - - - - - - distance (miles)

3 1/2 - - - - - - - - - - - - - - - - - - - 250

2 1/2 - - - - - - - - - - - - - - - - - - - - 175

Calculate the plane's altitude at 25miles from the airport

7/2 = 250

5/2 = 175

(7/2 - 5/2) miles = (250 - 175)miles

1 mile = 75 miles

From the above, the plane craft 's altitude reduces by 1 mile for every 75 miles moved

Therefore, at a distance of 25 mile from the airport,

Distance moved = (250 - 25)miles = 225miles

225miles / 75 miles = 3

Therefore, the altitude will reduce by 3 miles from the initial altitude at 250 miles

Therefore,

( 3 1/2 - 3) miles = 1/2 miles

Therefore, the plane craft will have an altitude of 1/2

.25

step-by-step explanation:

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