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Schach [20]
2 years ago
8

The ground temperature at an airport is 10 °C. The temperature decreases by 5.4 °C for every increase of 1 kilometer above the g

round. What is the temperature outside a plane flying at an altitude of 5 kilometers?
Mathematics
1 answer:
Paraphin [41]2 years ago
5 0

The temperature outside a plane flying at an altitude of 5 kilometers is -17°C.

Given that,

  • The ground temperature at the airport is 10 °C.
  • The decrease in temperature is 5.4°C for each & every rise in 1 kilometer.

Based on the above information, the temperature outside should be

The decrease in the temperature should be

= 5.4×5

= 27°C

Now the final temperature is

= 10 - 27

= -17°C

Therefore we can conclude that the temperature outside a plane flying at an altitude of 5 kilometers is -17°C.

Learn more: brainly.com/question/20459283

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Answer:

−21

Step-by-step explanation:

=12s5t−6t4−6s−21

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If the theater sells out, 2254 people would have attended. With each ticket being 20 dollars, $45,080 would have been collected at the box office
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find the values of the six trigonometric functions for angle theta in standard position if a point with the coordinates (1, -8)
frutty [35]

Answer:

cosФ = \frac{1}{\sqrt{65}} , sinФ = -\frac{8}{\sqrt{65}} , tanФ = -8, secФ = \sqrt{65} , cscФ = -\frac{\sqrt{65}}{8} , cotФ = -\frac{1}{8}

Step-by-step explanation:

If a point (x, y) lies on the terminal side of angle Ф in standard position, then the six trigonometry functions are:

  1. cosФ = \frac{x}{r}
  2. sinФ = \frac{y}{r}
  3. tanФ = \frac{y}{x}
  4. secФ = \frac{r}{x}
  5. cscФ = \frac{r}{y}
  6. cotФ = \frac{x}{y}
  • Where r = \sqrt{x^{2}+y^{2} } (the length of the terminal side from the origin to point (x, y)
  • You should find the quadrant of (x, y) to adjust the sign of each function

∵ Point (1, -8) lies on the terminal side of angle Ф in standard position

∵ x is positive and y is negative

→ That means the point lies on the 4th quadrant

∴ Angle Ф is on the 4th quadrant

∵ In the 4th quadrant cosФ and secФ only have positive values

∴ sinФ, secФ, tanФ, and cotФ have negative values

→ let us find r

∵ r = \sqrt{x^{2}+y^{2} }

∵ x = 1 and y = -8

∴ r = \sqrt{x} \sqrt{(1)^{2}+(-8)^{2}}=\sqrt{1+64}=\sqrt{65}

→ Use the rules above to find the six trigonometric functions of Ф

∵ cosФ = \frac{x}{r}

∴ cosФ = \frac{1}{\sqrt{65}}

∵ sinФ = \frac{y}{r}

∴ sinФ = -\frac{8}{\sqrt{65}}

∵ tanФ = \frac{y}{x}

∴ tanФ = -\frac{8}{1} = -8

∵ secФ = \frac{r}{x}

∴ secФ = \frac{\sqrt{65}}{1} = \sqrt{65}

∵ cscФ = \frac{r}{y}

∴ cscФ = -\frac{\sqrt{65}}{8}

∵ cotФ = \frac{x}{y}

∴ cotФ = -\frac{1}{8}    

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Answer:

Step-by-step explanation:

d = 3/2

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a₁ = 35/2 - 19×3/2 = -11

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