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Usimov [2.4K]
3 years ago
12

A quadratic equation is shown below:

Mathematics
1 answer:
AfilCa [17]3 years ago
5 0

the solutions to a general quadratic equation is

X=-b±√b²-4ac/2a  ,, When ax²+bx+c=0

the discriminant is the expression under the radical b²−4ac

Part A:

discriminant is (-16)² - 4(9)(60) = -1904

there are two complex solutions  

Part B:  

4x² + 8x − 5 = 0

4x² + 10x -2x - 5=0

2x ( 2x + 5) - 1(2 x + 5) =0  

(2x + 5)(2x-1) = 0

      /                     \

    /                          \

2x+5 = 0                 2x-1 =0    

x = -5/2                     x = 1/2

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3. At noon Lan is in his blue mini-van 300 km north of Makenna in her Bugatti. Lan heads south
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Answer:

The rate at which the distance between them is changing at 2:00 p.m. is approximately 1.92 km/h

Step-by-step explanation:

At noon the location of Lan = 300 km north of Makenna

Lan's direction = South

Lan's speed = 60 km/h

Makenna's direction and speed = West at 75 km/h

The distance  Lan has traveled at 2:00 PM = 2 h × 60 km/h = 120 km

The distance north between Lan and Makenna at 2:00 p.m = 300 km - 120 km = 180 km

The distance West Makenna has traveled at 2:00 p.m. = 2 h × 75 km/h = 150 km

Let 's' represent the distance between them, let 'y' represent the Lan's position north of Makenna at 2:00 p.m., and let 'x' represent Makenna's position west from Lan at 2:00 p.m.

By Pythagoras' theorem, we have;

s² = x² + y²

The distance between them at 2:00 p.m. s = √(180² + 150²) = 30·√61

ds²/dt = dx²/dt + dy²/dt

2·s·ds/dt = 2·x·dx/dt + 2·y·dy/dt

2×30·√61 × ds/dt = 2×150×75 + 2×180×(-60) = 900

ds/dt = 900/(2×30·√61) ≈ 1.92

The rate at which the distance between them is changing at 2:00 p.m. ds/dt ≈ 1.92 km/h

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