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True [87]
11 months ago
11

The length of a rectangle is 6 yd more than twice the width x. The area is 416 yd2. Find the dimensions of the rectangle.

Mathematics
1 answer:
kati45 [8]11 months ago
8 0

Answer:

13 yds * 32 yds

Step-by-step explanation:

A=area, l=length, w=width

A=w*l

l=6+2w

A=w*(6+2w)

A=6w+2w^2

416=6w+2w^2

416/2=(6w+2w^2)/2

208=3w+w^2

208-208=3w+w^2-208

w^2+3w-208=0

w^2+16w-13w-208=0

w(w+16)-13(w+16)=0

(w-13)(w+16)=0 ==> w+16=0 ==> w=-16, w can't be negative.

w-13=0

<u>w=13</u>

l=6+2w

l=6+2(13)

l=6+26

<u>l=32</u>

Dimensions: 13 * 32

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Answer:

(a) The critical number of f(x) are x=-4, 1

(b)

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Step-by-step explanation:

(a) The critical numbers of a function are given by finding the roots of the first derivative of the function or the values where the first derivative does not exist. Since the function is a polynomial, its domain and the domain of its derivatives is (-\infty, \infty). Thus:

\frac{df(x)}{dx}  = \frac{d(2x^3+9x^2-24x)}{dx} =6 x^2+18x -24\\6 x^2+18x -24=0\\\boxed{x=-4, x=1}

(b)

  • A function f(x) defined on an interval is monotone increasing on (a, b) if for every x_1, x_2 \in (a, b): x_1 implies f(x_1)
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Combining  the domain (-\infty, \infty) with the critical numbers we have the intervals (-\infty, -4), (-4, 1) and (1, \infty). Note that any of the points are included, in the case of the infinity it is by definition and the critical number are never included because the function monotony is not defined in the critical points, i.e. it is not monotone increasing or decreasing. Now, let's check for the monotony in each interval, for this, we check for the sign of the first derivative in each interval. Evaluating in each interval the first derivative (one point is enough), we obtain the monotony of the function to be:

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3 years ago
a. Verify that the given point lies on the curve. b. Determine an equation of the line tangent to the curve at the given point.
xxMikexx [17]

Answer:

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Step-by-step explanation:

Given:

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                               44 = 5x^2 + 3xy + 3y^2

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b. Determine an equation of the line tangent to the curve at the given point.

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Where, m is the gradient of the line.

            C is the y-intercept.

                              m = Δy / Δx = dy/dx

- We will take the derivative of the given curve with respect to x as follows:

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- To evaluate C, we will use the point (2,2) for linear expression above with m as follows:

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- The equation of the tangent is as follows:

                            y = 13*( -x/9 + 1/5)  

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