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jeyben [28]
3 years ago
9

2. y = k2 - 13k +42​

Mathematics
1 answer:
bekas [8.4K]3 years ago
6 0

For this case we must solve the following quadratic equation:

y = k ^ 2-13k + 42

With y = 0 we have:

k ^ 2-13k + 42 = 0

The roots will be given by:

k = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}

Where:

a = 1\\b = -13\\c = 42

Substituting:

k = \frac {- (- 13) \pm \sqrt {(- 13) ^ 2-4 (1) (42)}} {2 (1)}\\k = \frac {13 \pm \sqrt {169-168}} {2}\\k = \frac {13 \pm \sqrt {1}} {2}\\k = \frac {13 \pm1} {2}

Thus, we have two roots:

k_ {1} = \frac {13 + 1} {2} = \frac {14} {2} = 7\\k_ {2} = \frac {13-1} {2} = \frac {12} {2} = 6

Answer:

k_ {1} = 7\\k_ {2} = 6

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3 years ago
A lighthouse is built on an exposed reef, 4.5 miles off-shore. The shoreline is perfectly straight, and a town is located 6 mile
Alisiya [41]

Answer: 3.94 hours

Step-by-step explanation:

From the question, the following parameters are given:

Distance offshore = 4.5 miles

Distance downshore = 6 miles

Speed when running = 3.4 mph

Speed with boat = 1.8 mph

distance of boat rowing = sqrt(4.5^2 + x^2)

Where

Speed = distance/time

Time = distance/speed

Time = distance of boat rowing/1.8

distance of running = 6 - x

Time = (6-x)/3.4

total travel time

t = sqrt(4.5^2 + x^2)/1.8 + (6 - x)/3.4

dt/dx = x/(1.8×sqrt(4.5^2 + x^2) - 1/3.4

d^2t/dx^2 = +ve at any x

x is at its minimum when dt/dx=0

x/(1.8×sqrt(4.5^2 + x^2) = 1/3.4

x = 6/sqrt(253) × 4.5 = 1.697 miles

Substitute x in

t = sqrt(4.5^2 + x^2)/1.8 + (6 - x)/3.4

We obtaine

t = sqrt(4.5^2 + 1.697^2)/1.8 + (6 - 1.69)/3.4

t = 2.672+ 1.266 = 3.94 hours

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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