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enot [183]
3 years ago
8

It took Gustavo 5 hours to drive to a job interview. On the way home, he was able to increase his average speed by 20mph and mak

e the return drive in only 3 hours. Find his average speed on the return drive.
Mathematics
1 answer:
Alik [6]3 years ago
8 0
<h2>Answer:</h2>

His average speed on the return drive is:

                             50 mph

<h2>Step-by-step explanation:</h2>

Let d be the distance from home to the place of his job interview.

Let s be the average speed when he drive to his job interview.

Now, it is given that:

It took Gustavo 5 hours to drive to a job interview.

We know that distance is the product of speed and time.

i.e.

Distance=Speed\times Time

Hence, we have:

d=5s--------(1)

Also,

On the way home, he was able to increase his average speed by 20 mph and make the return drive in only 3 hours.

His average speed back home is: s+20

and his time is: 3 hours

Hence, we have:

d=3(s+20)\\\\i.e.\\\\d=3s+60----------(2)

From equation (1) and equation (2) we have:

5s=3s+60\\\\i.e.\\\\5s-3s=60\\\\2s=60\\\\s=\dfrac{60}{2}\\\\s=30\ mph

Hence, his average speed on the return drive is: s+20

                                                                              = 30+20

                                                                             = 50 mph

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Let (x_1,y_1)=(2,4)

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The point P(7, −2) lies on the curve y = 2/(6 − x). (a) If Q is the point (x, 2/(6 − x)), use your calculator to find the slope
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Answer:

a) (i) m = 2.22, (ii) m = 2, (iii) m = 2, (iv) m = 2, (v) m = 1.82, (vi) m = 2, (vii) m = 2, (viii) m = 2; b) m \approx 2; c) The equation of the tangent line to curve at P (7, -2) is y = 2\cdot x + 12.

Step-by-step explanation:

a) The slope of the secant line PQ is represented by the following definition of slope:

m = \frac{\Delta y}{\Delta x} = \frac{y_{Q}-y_{P}}{x_{Q}-x_{P}}

(i) x_{Q} = 6.9:

y_{Q} =\frac{2}{6-6.9}

y_{Q} = -2.222

m = \frac{-2.222 + 2}{6.9-7}

m = 2.22

(ii) x_{Q} = 6.99

y_{Q} =\frac{2}{6-6.99}

y_{Q} = -2.020

m = \frac{-2.020 + 2}{6.99-7}

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(iii) x_{Q} = 6.999

y_{Q} =\frac{2}{6-6.999}

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m = 2

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y_{Q} = -2.0002

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m = \frac{-1.9998 + 2}{7.0001-7}

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m \approx \frac{f(6.9999)+f(7.0001)}{2}

m \approx \frac{2+2}{2}

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The slope of the tangent line to the curve at P(7, -2) is 2.

c) The equation of the tangent line is a first-order polynomial with the following characteristics:

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Where:

x - Independent variable.

y - Depedent variable.

m - Slope.

b - x-Intercept.

The slope was found in point (b) (m = 2). Besides, the point of tangency (7,-2) is known and value of x-Intercept can be obtained after clearing the respective variable:

-2 = 2 \cdot 7 + b

b = -2 + 14

b = 12

The equation of the tangent line to curve at P (7, -2) is y = 2\cdot x + 12.

7 0
3 years ago
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