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tangare [24]
3 years ago
10

An athlete is running in a marathon that eventually finishes at a local high school. After x minutes, the distance the athlete i

s from the school is given by y (in miles), where x + 10y = 260 (Note: 0 ≤ x ≤ 10 and 0 ≤ y ≤ 26). Complete the steps below to understand how to find the intercepts of the equation and what they represent.
Use the relationship between x and y to find the time it takes for the runner to reach the school.
Mathematics
2 answers:
Ilya [14]3 years ago
8 0
Given <span>x + 10y = 260 ; talso 0 ≤ x ≤ 10 and 0 ≤ y ≤ 26

to find the intercepts, set x=0

0 + 10(y-intercept) = 260

y-intercept = 26 miles at x=0; so runner is 26 miles from school at the start


then set y=0

x-intercept + 10(0) = 260

x-intercept = 260 minutes at y=0; so runner reaches school after 260 minutes

NOTE: the x-intercept is beyond the given range of </span>0 ≤ x ≤ 10; perhaps there is a typo n the range should <span>0 ≤ x ≤ 1000 instead?</span><span>


</span>
RoseWind [281]3 years ago
3 0
Here, we have the limits: <span> 0 ≤ x ≤ 10 and 0 ≤ y ≤ 26

When, x = 0, y = 260/10 = 26
x = 1, y = 250/10 = 25.9
x = 2, y = 258/10 = 25.8
x = 3, y = 257/10 = 25.7

and so on....for x = 10, y = 250/10 = 25

Intercepts would be: (0, 26), (1, 25.9) (2, 25.8) (3, 25.7), (4, 25.6), (5, 25.5) (6, 25.4) (7, 25.3) (8, 25.2), (9, 25.1), (10, 25)

x-coordinate represents the time taken by the athlete, and y represents the distance, he covered in that time.

Hope this helps!</span>
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Answer:

\mathrm{Parabola\:focus\:given}\:x^2=-20y:\quad \left(0,\:-5\right)

Step-by-step explanation:

Given the equation

x2 = -20y

A parabola is the locus of points such that the distance to a point the focus equals the distance to a line the directrix.

4p\left(y-k\right)=\left(x-h\right)^2 is the standard equation for an up-down facing parabola with vertex at (h, k), and a focal length |p|.

so

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\mathrm{Switch\:sides}

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\mathrm{Factor\:}4

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4\left(-5\right)y=x^2

\mathrm{Rewrite\:as}

4\left(-5\right)\left(y-0\right)=\left(x-0\right)^2

\left(h,\:k\right)=\left(0,\:0\right),\:p=-5

Parabola is symmetric around the y-axis and so the focus lies a distance\ p from the center (0, 0) along the y-axis.

\left(0,\:0+p\right)

=\left(0,\:0+\left(-5\right)\right)

\mathrm{Refine}

\left(0,\:-5\right)

Therefore,

\mathrm{Parabola\:focus\:given}\:x^2=-20y:\quad \left(0,\:-5\right)

Please check the attached figure too.

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