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.
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Answer:
(-3+
) /2, (3+) / 2 which is approx -3.791 and 0.791
Step-by-step explanation:
i used the quadratic formula: [-b± sq rt (b² - 4ac) / 2a]
in this problem, a = 1, b = 3, and c = -3
First: Slope of -x+2y=42y=x+4y=x/2+2So, m=1/2.
Second: Slope of the perpendicular line: mp=-2.
Third: Find the line with slope -2 and passes through the point (−2, 1)y-y1=m(x-x1)y-1=-2(x+2)y=-2x-4+1y=-2x-3
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Answer:
8.5
Step-by-step explanation:
4 + (6)(6)/8=
17/2 = 8.5
Answer:
y=3x/6+x
Step-by-step explanation:
