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Bond [772]
3 years ago
5

The ground temperature at an airport is 10 °C. The temperature decreases by 6 °C for every increase of 1 kilometer above the gro

und. What is the temperature outside a plane flying at an altitude of 5 kilometers?
Mathematics
1 answer:
lions [1.4K]3 years ago
3 0

Answer:

degrees Celsius.

Step-by-step explanation:

Let x represent number of kilometers.

We have been given that the temperature decreases by 6.8 degrees Celsius for every increase of 1 kilometer above the ground. So temperature decreased in x hours would be .

The ground temperature at Brigham Airport is 12 degrees Celsius. The temperature at an altitude of x kilometers would be .

To find temperature outside a plane flying at an altitude of 5 kilometers, we will substitute  in our expression as:

Therefore, the temperature outside a plane flying at an altitude of 5 kilometers would be  degrees Celsius.

I hope this helps, this is close I hope!

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PLEASE HELP FAST!!
Mice21 [21]

Answer:

  • 367.6 m

Step-by-step explanation:

Use trigonometry

<u>Let the height of the cloud be x, then we have:</u>

  • tan 35° = x/525
  • x = 525 tan 35°
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The height of the cloud layer is 367.6 m

8 0
3 years ago
Complete the solution of the equation. Find the value of y when x equals 15.
GuDViN [60]
So first, we want to substitute in 15 to all the x variables.

So,

3x - 8y = 5

Becomes

3 * 15 - 8y = 5

Or, 

45 - 8y = 5

So here, you can subtract 45 from both sides to move all like terms to one side.

-8y = 5 - 45

And then simplify:

-8y = -40

Then divide both sides by -8

y = 5
4 0
3 years ago
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Whats the surface area
Lena [83]

Answer:

20 units^2

Step-by-step explanation:

5 x 4

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3 0
3 years ago
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Find the x- and y-intercepts of the following equation:<br> 2x+3y=-24
Ilya [14]

Answer:

Algebra 

 Topics 

How do you find the intercepts of x2y−x2+4y=0?

Algebra  Graphs of Linear Equations and Functions  Intercepts by Substitution

2 Answers

Gió

Mar 24, 2015

For the intercepts you set alternately x=0 and y=0 in your function:

and graphically:

Answer link

Alan P.

Mar 24, 2015

On the X-axis y=0

So

x2y−x2+4y=0

becomes

x2(0)−x2+4(0)=0

→−x2=0

→x=0

On the Y-axis x=0

and the original equation

x2y−x2+4y=0

becomes

(0)2y−(0)2+4y=0

→y=0

The only intercept for the given equation occurs at (0,0)

Answer link

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4 0
2 years ago
How many solutions does the equation have? 0 = -8q + 8q
lawyer [7]
Zeroooo solutions bc i did it earlier
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3 years ago
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