The longest possible integer length of the third side of the triangle is 6 < x < 28
The sum of any two sides must be greater than the third side for a triangle to exist
let the third side be x
x + 11 > 17 and x + 17 > 11 and 11 + 17 > x
x > 6 and x > - 6 and x < 28
The longest possible integer length of the third side of the triangle is 6 < x < 28
The length of the 3 sides of a triangle needs to always be among (however no longer the same) the sum and the difference of the opposite two sides. As an example, take the instance of two, 6, and seven. and. consequently, the third side period should be extra than 4 and less than 8.
Learn more about triangles here: brainly.com/question/1675117
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Answer:720
Explanation:
Volume= lwh, 8•9•10= 720
<h3>
Answers:</h3>
- a) HR = 24
- b) AT = 32
- c) MR = 29
- d) R is the midpoint of <u> MH </u> and <u> AT </u>
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Explanation:
a)
MR = 24 is given. Note the single tickmark on this segment. The other segment HR has the same tickmark, so it is the same length as MR.
HR = MR = 24
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b)
The double tickmarks tell us TR and RA are the same length. They combine to form segment AT. So this means segment AT is cut in half to get TR and RA.
TR = (1/2)*(AT)
2*TR = AT.... multiply both sides by 2
AT = 2*TR
AT = 2*16
AT = 32
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c)
MH is cut in half at point R for similar reasoning as part b)
Put another way, MH is twice as long as MR.
MH = 2*MR
58 = 2*MR
2*MR = 58
MR = 58/2
MR = 29
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d)
R is the midpoint of segment MH because R cuts MH in half. Point R bisects MH.
R is also the midpoint of AT for similar reasoning. It's all based on the tickmarks telling us which segments are congruent to one another.
Answer:
C
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5y+2 i think because those are parallel lines:)