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Dahasolnce [82]
3 years ago
12

The wind-chill index is modeled by the function below where T is the temperature (°C) and v is the wind speed (km/h).

Mathematics
1 answer:
SpyIntel [72]3 years ago
6 0
When T= 13°C and v = 34 km/h
W = 13.12 + 0.6215(13) - 11.37(34)^0.16 + 0.3965(13)(34)^0.16 = 13.12 + 8.0795 - 11.37(1.7581) + 5.1545(1.7581) = 21.1995 - 19.9893 + 9.0620 = 10.27°C.

If T decrease by 1°C (T = 12°C), then
W = 13.12 + 0.6215(12) - 11.37(34)^0.16 + 0.3965(12)(34)^0.16 = 8.95°C
Therefore, W decreased by 10.27°C - 8.95°C = 1.3°C

If v increases by 1 km/h (v = 35 km/h), then
W = 13.12 + 0.6215(13) - 11.37(35)^0.16 + 0.3965(13)(35)^0.16 = 10.22°C
Therefore, W decreaseb by 10.27°C - 10.22°C = 0.05°C

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i hope it helps

4 0
3 years ago
ABC IS a right angled triangle at B and D is a point on BC. If AD=18cm , BD=9cm and CD = 4 cm, find AC
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Explain why the oppisite of a positive number will always be a negitive number​
Aloiza [94]

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Step-by-step explanation:

7 0
3 years ago
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Simplify: 18q2−45q+25/9q2−25.
Svet_ta [14]

Answer:

\frac{(6q-5)}{(3q+5)}

Step-by-step explanation:

Given: \frac{18q^{2}-45q+25 }{9q^{2}-25 }

Factorizing the numerator and the denominator, we have:

18q^{2} - 45q + 25 = 18q^{2} - 30q - 15q + 25

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and

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So that,

\frac{18q^{2}-45q+25 }{9q^{2}-25 } = \frac{(3q-5)(6q-5)}{(3q-5)(3q+5)}

                 = \frac{(6q-5)}{(3q+5)}

Therefore,

\frac{18q^{2}-45q+25 }{9q^{2}-25 } = \frac{(6q-5)}{(3q+5)}

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3 years ago
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