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vazorg [7]
3 years ago
15

(a) find the slope m of the tangent to the curve y = 5/ x at the point where x = a > 0.(b) find equations of the tangent line

s at the points (1, 5) and 4, 5 2 .
Mathematics
1 answer:
Trava [24]3 years ago
4 0
<span>(a) the slope of curve is calculated from the derivative of the curve expression y=5/x. In this problem, the slope m=-5/(a^2). (b) x=1, y=5,and m=-5, the tangent line is y-5=-5*(x-1). x=4, y=5/4, and m=-5/16, the tangent line is y-5/4=-5/16*(x-4)</span>
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A motel rents double rooms at $34 per day and single rooms at $26 per day. If 26 rooms were rented one day for a total of $804,
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Answer: 16 double rooms and 10 single rooms were rented.

Step-by-step explanation:

Let x represent the number of double rooms that were rented.

Let y represent the number of single rooms that were rented.

The total number of rooms rented in a day is 26. It means that

x + y = 26

A motel rents double rooms at $34 per day and single rooms at $26 per day. If all the rooms that were rented for one day cost a total of $804, it means that

34x + 26y = 804 - - - - - - - - - - -1

Substituting x = 26 - y into equation 1, it becomes

34(26 - y) + 26y = 804

884 - 34y + 26y = 804

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Write the equation of the line that is the perpendicular bisector of the segment with endpoints (4, 1) and (2, -5)
Sedbober [7]

Answer:   \bold{y=-\dfrac{1}{3}x-1}

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

(4, 1) & (2, -5)

First, find the slope (m) and then the perpendicular (opposite reciprocal) slope:

m=\dfrac{y_2-y_1}{x_2-x_1}\\\\\\m=\dfrac{-5-1}{2-4} = \dfrac{-6}{-2}=3\quad \rightarrow \quad m_{\perp}=-\dfrac{1}{3}\\

Next, find the midpoint of (4, 1) and (2, -5):

Midpoint=\bigg(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\bigg)\\\\\\.\qquad \qquad=\bigg(\dfrac{4+2}{2},\dfrac{1-5}{2}\bigg)\\\\\\.\qquad \qquad=\bigg(\dfrac{6}{2},\dfrac{-4}{2}\bigg)\\\\\\.\qquad \qquad=(3, -2)

Lastly, input the perpendicular slope and the midpoint into the Point-Slope formula to find the equation of the line:

y - y_1 = m_{\perp}(x - x_1)\\\\y - (-2) = -\dfrac{1}{3}(x - 3)\\\\y + 2=-\dfrac{1}{3}x +1\\\\y =-\dfrac{1}{3}x - 1\\


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