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Fiesta28 [93]
3 years ago
10

A piece of wire measuring 20 feet is attached to a telephone pole as a guy wire. The distance along the ground from the bottom o

f the pole to the end of the wire is 4 feet greater than the height where the wire is attached to the pole how far up the pole does the guy wire reach? PLEASE SHOW ALL WORK !!!!

Mathematics
1 answer:
dem82 [27]3 years ago
8 0

Answer: Height at which the wire is attached to the pole is 12 feet.

Explanation:

Since we have given that

Length of the wire = 20 feet

Let the height at which wire is attached to the pole be h

and distance along the ground from the bottom of the pole to the end of the wire be x+4

Now, it forms  a right angle triangle so, we can apply "Pythagorus theorem".

H^2=B^2+P^2\\\\20^2=x^2+(x+4)^2\\\\400=x^2+x^2+8x+16\\\\400=2x^2+8x+16\\\\400=2(x^2+4x+8)\\\\200=x^2+4x+8\\\\x^2+4x-192=0\\\\x^2+16x-12x-192=0\\\\x(x+16)-12(x+16)=0\\\\(x+16)(x-12)=0\\\\x=-16\ and\ 12

But height cant be negative so, height will be 12 feet.

Hence, height at which the wire is attached to the pole is 12 feet.


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NOTES:

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  2. Plug the x-value from 1 (above) into the given equation to find the y-value. <em>(which is the max/min)</em>
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Rate of Change: slope between the two given points.

********************************************************************************************

1. f(x) = (x-1)²(x + 6)

a) Degree = 3

b) end behavior:

  • Coefficient is positive so right side goes to positive infinity
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c) (x - 1)²(x + 6) = 0

   x - 1 = 0                     x + 6 = 0

       x = 1 (M=2)                   x = -6 (M=1)

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y = (-3.5 - 1)²(-3.5 + 6)

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e) see attachment #1

f) The interval at which the graph increases is: (-∞, -3.5)U(1, ∞)

g) The interval at which the graph decreases is: (-3.5, 1)

h) f(-1) = (-1 - 1)²(-1 + 6)

          = (-2)²(5)

          = 20

    f(0) = (0 - 1)²(0 + 6)

          = (-1)²(6)

          = 6

Find the slope between (-1, 20) and (0, 6)

m = \frac{20-6}{-1-0}

   = \frac{14}{-1}

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********************************************************************************************

2.    y = x³+3x²-10x

         = x(x² + 3x - 10)      

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b) end behavior:

   Coefficient is positive so right side goes to positive infinity

   Degree is odd so left side goes to negative infinity

c) x(x + 5)(x - 2) = 0

   x = 0                     x + 5 = 0                     x - 2 = 0

   x = 0 (M=1)                   x = -5 (M=1)                x = 2 (M=1)

d) The midpoint between -5 and 0 is -2.5, so axis of symmetry is at x = -2.5

y = -2.5(-2.5 + 5)(-2.5 - 2)

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The midpoint between 0 and 2 is 1, so axis of symmetry is at x = 1

y = 1(1 + 5)(1 - 2)

  =  1(6)(-1)

  = -6

-6 is the relative min

e) see attachment #2

f) The interval at which the graph increases is: (-∞, -2.5)U(1, ∞)

g) The interval at which the graph decreases is: (-2.5, 1)

h) f(-1) = -1(-1 + 5)(-1 - 2)

********************************************************************************************

3. y = -x(x + 2)(x - 7)(x - 3)

a) Degree = 4

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   Coefficient is negative so right side goes to negative infinity

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  -x = 0                     x + 2 = 0                     x - 7 = 0             x - 3 = 0

   x = 0 (M=1)                 x = -2 (M=1)                x = 7 (M=1)          x = 3 (M=1)

d) The midpoint between -2 and 0 is -1, so axis of symmetry is at x = -1

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  =  1(1)(-8)(-4)

  = 32

32 is a relative max

The midpoint between 0 and 3 is 1.5, so axis of symmetry is at x = 1.5

y = -(1.5)(1.5 + 2)(1.5 - 7)(1.5 - 3)

  =  -1.5(3.5)(-5.5)(-1.5)

  = -43.3125

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140 is the relative max

e) see attachment #3

f) The interval at which the graph increases is: (-∞, -1)U(1.5, 5)

g) The interval at which the graph decreases is: (-1, 1.5)U(5, ∞)

h) f(-1) = -(-1)(-1 + 2)(-1 - 7)(-1 - 3)

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    f(0) = -(0)(0 + 2)(0 - 7)(0 - 3)

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Find the slope between (-1, 32) and (0, 0)

m = \frac{32-0}{-1-0}

   = \frac{32}{-1}

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