Answer:
Find the midsegment of the triangle which is parallel to CA.
Tip
- A midsegment of a triangle is a segment connecting the midpoints of two sides of a triangle.
- This segment has two special properties. It is always parallel to the third side, and the length of the midsegment is half the length of the third side.
- If two segments are congruent, then they have the same length or measure.In other words, congruent sides of a triangle have the same length.
We have to find the segment which is parallel to CA.
From the given data,
The segment EG is the midsegment of the triangle ABC.
So we have,
A midsegment of a triangle is a segment connecting the midpoints of two sides of a triangle. This segment has two special properties. It is always parallel to the third side.
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Answer:
2628
Step-by-step explanation:
7/4 = 1.75 = 175%
x is the number of no votes
4599/x = 7/4
x = 4599/(7/4)
x = 2628
Answer:
x<8 or x>13
Step-by-step explanation:
x-5 < 3
x < 3 + 5
x <8
OR
x – 5 > 8
x > 8 +5
x >13
combining the two, the combined answer for x is
x<8 or x>13
Answer:
83.33%
Step-by-step explanation:
300 - 50 = 250
250/300 = 0.8333 or 83.33%
Answer:
25.133 units
Step-by-step explanation:
Since the density ρ = r, our mass is
m = ∫∫∫r³sinθdΦdrdθ. We integrate from θ = 0 to π (since it is a hemisphere), Φ = 0 to 2π and r = 0 to 2 and the maximum values of r = 2 in those directions. So
m =∫∫[∫r³sinθdΦ]drdθ
m = ∫[∫2πr³sinθdθ]dr ∫dФ = 2π
m = ∫2πr³∫sinθdθ]dr
m = 2π∫r³dr ∫sinθdθ = 1
m = 2π × 4 ∫r³dr = 4
m = 8π units
m = 25.133 units