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r-ruslan [8.4K]
4 years ago
10

An airplane has begun its descent for a landing. When the airplane is 150 miles west of its destination, its altitude is 25,000

feet. When the airplane is 90 miles west of its destination, its altitude is 19,000 feet. If the airplane's descent is modeled by a linear function, where will the airplane be in relation to the runway when it hits ground level?
Mathematics
2 answers:
Anastasy [175]4 years ago
7 0
We have two points on a plane p1(150, 25000), p2(90, 19000) and we can find a linear equation from them, lets calculate the slope:
m = (y2 - y1)/(x2 - x1)
where x1, y1 and x2, y2 are the point's coordinates:
m = (19000 - 25000)/(90 - 150)
m = -6000/-60
m = 100
so we have the slope now, and we can use the equation of the line for a point and having the slope, that equation is:
y - y1 = m(x - x1)
so we substitute:
y - 25000 = 100(x - 150)
y = 100x + 25000 - 15000
y = 100x + 10000
so this is the linear equation that models the airplane descent, when the airplane hits the ground, then y = 0, and we need to find the x, that is the position in relation to the runway:
<span>0 = 100x + 10000
</span>-100x = 10000
x = -100
therefore, when the airplane hits the ground, y = 0, the position in relation to the runway will be -100, that is 100 miles to the east of the runway
kvv77 [185]4 years ago
4 0

Answer:

My user name is Kim Namjoon but I don't got IQ 148 ok

Step-by-step explanation:

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\large\boxed{18\%}

Step-by-step explanation:

\$21.73-\text{selling price}\\\\\$26.50-\text{starting price}\\\\\$26.50-\$21.73=\$4.77-\text{difference of price - discount}\\\\METHOD\ 1:\\\\\dfrac{4.77}{26.50}=0.18=0.18\cdot100\%=\boxed{18\%}\\\\METHOD\ 2:\\\\\begin{array}{ccc}\$26.50&-&100\%\\\\\$4.77&-&p\%\end{array}\qquad\text{cross multiply}\\\\\\26.50p=(4.77)(100)\\\\26.50p=477\qquad\text{divide both sides by 26.50}\\\\\boxed{p=18\ (\%)}

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