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jok3333 [9.3K]
3 years ago
15

James is fencing off a triangular region for his cats to play in. From the first fence post, he walks 13 meters in a straight li

ne and plants the second fence post. From there, James heads in another direction for 5 meters and places the third fence post. If the distance between the first and third fence posts is 9 meters, find the area of the enclosed region. Round the answer to the nearest tenth.
Mathematics
1 answer:
GenaCL600 [577]3 years ago
4 0
You can use Heron's formula to do this  

Area of triangle = sqrt[ s(s-a)(s-b)(s-c) ]   where s = half the perimeter of the triangle = 
1/2 * 13+5+9 = 13.5
so required area =  sqrt [ (13.5)(13.5-13)(13.5-5)(13.5-9)]
                            =  16.1  square meters
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Find the roots of h(t) = (139kt)^2 − 69t + 80
Sonbull [250]

Answer:

The positive value of k will result in exactly one real root is approximately 0.028.

Step-by-step explanation:

Let h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80, roots are those values of t so that h(t) = 0. That is:

19321\cdot k^{2}\cdot t^{2}-69\cdot t + 80=0 (1)

Roots are determined analytically by the Quadratic Formula:

t = \frac{69\pm \sqrt{4761-6182720\cdot k^{2} }}{38642}

t = \frac{69}{38642} \pm \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321}  }

The smaller root is t = \frac{69}{38642} - \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321}  }, and the larger root is t = \frac{69}{38642} + \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321}  }.

h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80 has one real root when \frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} = 0. Then, we solve the discriminant for k:

\frac{80\cdot k^{2}}{19321} = \frac{4761}{1493204164}

k \approx \pm 0.028

The positive value of k will result in exactly one real root is approximately 0.028.

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MissTica

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Answer:

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Step-by-step explanation:

<em>g(x) = </em>\sqrt{x+2}<em> - 3 </em>

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Step-by-step explanation:

(25/50) x 75 = 37.5

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