The two lines are parallel, therefor 90-28=internal angle=62. Using trigonometry, Cosine=a/h. Cos(62)=350/x. Rearrange to get 350/Cos(62) = 519.7(1dp)
It has been proven that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
<h3>How to prove a Line Segment?</h3>
We know that in a triangle if one angle is 90 degrees, then the other angles have to be acute.
Let us take a line l and from point P as shown in the attached file, that is, not on line l, draw two line segments PN and PM. Let PN be perpendicular to line l and PM is drawn at some other angle.
In ΔPNM, ∠N = 90°
∠P + ∠N + ∠M = 180° (Angle sum property of a triangle)
∠P + ∠M = 90°
Clearly, ∠M is an acute angle.
Thus; ∠M < ∠N
PN < PM (The side opposite to the smaller angle is smaller)
Similarly, by drawing different line segments from P to l, it can be proved that PN is smaller in comparison to all of them. Therefore, it can be observed that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
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Answer: 96 ft²
Step-by-step explanation:
<u>This figure can be divided into two rectangles:</u>
- The first rectangle has dimensions 6 ft by 14 ft.
- The second rectangle had dimensions 4 ft by 3 ft.
<u />
<u>Find the area of both rectangles:</u>
- The first rectangle: 6 × 14 = 84 ft²
- The second rectangle: 4 × 3 = 12 ft²
<u />
<u>Add them together to find the total area of the figure:</u>
84 + 12 = 96 ft²
Answer:
i would say that i would take 5 days
Step-by-step explanation:
9(n+3)=7n-3
1) distribute 9 into (n+3)
9n+27=7n-3
2) subtract 7n on both sides
2n+27=-3
3) subtract 27 on both sides
2n=-30
4) divide by 2
n=-15