Answer: segment KH
Explanation:
The altitude of a triangle is defined as a segment connecting one of the vertex of the triangle with the opposite side of the triangle (or with an external prolongation of it), forming a right angle with it.
Note that the altitude of a triangle can also be outside the area of the triangle.
If we look at the picture, we see that:
- IH is a side, so it is not an altitude of the triangle
- KH is an altitude of the triangle, since it connects the vertex H with the external prolongation of IG, and it makes a right angle with it
- GJ: we don't know if it is an altitude, since we don't know if it forms a right angle or not
- GH: it is a side of the triangle, so it is not an altitude
So, the correct answer is segment KH.
I uploaded it here as the answers Bc
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Answer:
Jynessa wants to order these fractions: StartFraction 4 over 9 EndFraction, two-thirds, one-sixth, Negative 2 and one-half. What should she use as her common denominator? 6 9 12 18Jynessa wants to order these fractions: StartFraction 4 over 9 EndFraction, two-thirds, one-sixth, Negative 2 and one-half. What should she use as her common denominator? 6 9 12 18Jynessa wants to order these fractions: StartFraction 4 over 9 EndFraction, two-thirds, one-sixth, Negative 2 and one-half. What should she use as her common denominator? 6 9 12 18Jynessa wants to order these fractions: StartFraction 4 over 9 EndFraction, two-thirds, one-sixth, Negative 2 and one-half. What should she use as her common denominator? 6 9 12 18Jynessa wants to order these fractions: StartFraction 4 over 9 EndFraction, two-thirds, one-sixth, Negative 2 and one-half. What should she use as her common denominator? 6 9 12 18Jynessa wants to order these fractions: StartFraction 4 over 9 EndFraction, two-thirds, one-sixth, Negative 2 and one-half. What should she use as her common denominator? 6 9 12 18Jynessa wants to order these fractions: StartFraction 4 over 9 EndFraction, two-thirds, one-sixth, Negative 2 and one-half. What should she use as her common denominator? 6 9 12 18
Step-by-step explanation:
Answer:416
Step-by-step explanation: Based on the given problem above, it requires a trigonometric solution. So here it goes:
The climb distance is:200 * Sec [ArcTan [10/200]] = 10√ 401 km
and the decent distance is:300 * Sec [ArcTan [10/300]] = 10√901 km
So add the given results above with 500 km and this will be the additional distance that plane moves through the air.
The answer would be in the unit meters.
((10√401 +10√901−500)∗1000)=416 meters
Hope this is the answer that you are looking for.
