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castortr0y [4]
3 years ago
7

What is the slope of the line passing through the points (1, -5) and (4,1)

Mathematics
2 answers:
Hoochie [10]3 years ago
8 0
1-(-5). 6
———- = —- = 2
4-1 3
never [62]3 years ago
3 0

The slope of the line passing through (1,-5) and (4,1) is 2

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Since in the above case, the beaker has two sections each with different radius and height, we will divide this problem into two parts.

We will calculate the volume of both the beakers separately and then add them up together to get the volume of the beaker.

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3 years ago
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}

m = 2

(iii) x_{Q} = 6.999

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

y_{Q} = -2.002

m = \frac{-2.002 + 2}{6.999-7}

m = 2

(iv) x_{Q} = 6.9999

y_{Q} =\frac{2}{6-6.9999}

y_{Q} = -2.0002

m = \frac{-2.0002 + 2}{6.9999-7}

m = 2

(v) x_{Q} = 7.1

y_{Q} =\frac{2}{6-7.1}

y_{Q} = -1.818

m = \frac{-1.818 + 2}{7.1-7}

m = 1.82

(vi) x_{Q} = 7.01

y_{Q} =\frac{2}{6-7.01}

y_{Q} = -1.980

m = \frac{-1.980 + 2}{7.01-7}

m = 2

(vii) x_{Q} = 7.001

y_{Q} =\frac{2}{6-7.001}

y_{Q} = -1.998

m = \frac{-1.998 + 2}{7.001-7}

m = 2

(viii)  x_{Q} = 7.0001

y_{Q} =\frac{2}{6-7.0001}

y_{Q} = -1.9998

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

m = 2

b) The slope at P (7,-2) can be estimated by using the following average:

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

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

m \approx 2

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:

y = m\cdot x + b

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
2 years ago
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