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slega [8]
3 years ago
5

Jason is laying out a square garden on his property. The space measures 46 feet diagonally. How long is each side of the garden?

Mathematics
1 answer:
ryzh [129]3 years ago
3 0
46 ft all around the square
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What is the solution to the following equation x^2+5x+7=0
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Find the area and perimeter of that triangle​
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Find the general expression for the slope of a line tangent to the curve of y=2x^2+4x at the point P(x,y) . Then find the slopes
ArbitrLikvidat [17]

Complete Question

Find the general expression for the slope of a line tangent to the curve of y=2x^2+4x at the point P(x,y) . Then find the slopes for x = 3 and x=0.5. Sketch the curve and the tangent lines. What is the general expression for the slope of a line tangent to the curve of the function y=2x^2+4 at the point P(x,y) ​?

Answer:

The  generally expression for the slope of y  = 2x^2 + 4x is  y' =  4x +4

The graph is shown on the first uploaded image

The  generally expression for the slope of y=2x^2+4 is   y' =  4x

Step-by-step explanation:

From the question we are told that

  The  equation of the curve is y  = 2x^2 + 4x

First we differentiate the equation

So  

     y' =  4x +4

Therefore the generally expression for the slope tangent to the curve y=2x^2+4x is   y' =  4x +4

The  next step is to substitute for x =  3 and  x =  0.5

So  for x_1 =  3

    y' =  4(3) +4

     y' =m_1=  16

And  for  x_2 =  0.5

      y' =  4(0.5) +4

       y' =m_2=  6

Here m_1  and  m_2 are slops of the curve

Next we obtain the coordinates of the tangent lines

So  at x_1 =  3

   y_1  = 2(3)^2 + 4(3)

  y_1  =  21

So the coordinate for the first tangent line is  

    (x_1 , y_1 ) =  (3 ,  21)

At  x_2 = 0.5      

    y_2  = 2(0.5)^2 + 4(0.5)

=>  y_2  = 2.5

So the coordinate for the second  tangent line is  

    (x_2 , y_2 ) =  (0.5 ,  2.5)

Next we obtain the equation for the tangent lines

 So generally the slope is mathematically represented as

        m  =  \frac{y - y_1 }{x-x_1}

For   (x_1 , y_1 ) =  (-3 ,  21) and  y' =m_1=  16

       16 =  \frac{y -21 }{x-3)}

=>    y   = 16x - 27

For  (x_2 , y_2 ) =  (0.5 ,  2.5) and  y' =m_2=  6

       6  =  \frac{y -2.5 }{x-0.5}

       y  = 6x -0.5

Generally the general expression for the slope of a line tangent to the curve of the function y=2x^2+4 at the point P(x,y) is mathematically evaluated by differentiating  y=2x^2+4 as follows

     y' =  4x

     

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