Proof
Provide the missing reasons for the proof of part of the triangle midsegment theorem.
Given: K is the midpoint of MJ.
L is the midpoint of NJ.
Prove: MN = 2KL
The complete answer is attached in the diagram below.
The complete answer for the missing reasons is attached below in the diagram.
Please check the figure.
Keywords: statement, proof, reason
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Hello,
I note (a,b,c) the result of a quarters, b dimes and c pennies:
2 solutions:
106=( 3, 3, 1)=( 1, 8, 1)
106=( 0, 0, 106) but : 100= 0*25+ 0*10+ 100
106=( 0, 1, 96) but : 100= 0*25+ 1*10+ 90
106=( 0, 2, 86) but : 100= 0*25+ 2*10+ 80
106=( 0, 3, 76) but : 100= 0*25+ 3*10+ 70
106=( 0, 4, 66) but : 100= 0*25+ 4*10+ 60
106=( 0, 5, 56) but : 100= 0*25+ 5*10+ 50
106=( 0, 6, 46) but : 100= 0*25+ 6*10+ 40
106=( 0, 7, 36) but : 100= 0*25+ 7*10+ 30
106=( 0, 8, 26) but : 100= 0*25+ 8*10+ 20
106=( 0, 9, 16) but : 100= 0*25+ 9*10+ 10
106=( 0, 10, 6) but : 100= 0*25+ 10*10+ 0
106=( 1, 0, 81) but : 100= 1*25+ 0*10+ 75
106=( 1, 1, 71) but : 100= 1*25+ 1*10+ 65
106=( 1, 2, 61) but : 100= 1*25+ 2*10+ 55
106=( 1, 3, 51) but : 100= 1*25+ 3*10+ 45
106=( 1, 4, 41) but : 100= 1*25+ 4*10+ 35
106=( 1, 5, 31) but : 100= 1*25+ 5*10+ 25
106=( 1, 6, 21) but : 100= 1*25+ 6*10+ 15
106=( 1, 7, 11) but : 100= 1*25+ 7*10+ 5
106=( 1, 8, 1) is good
106=( 2, 0, 56) but : 100= 2*25+ 0*10+ 50
106=( 2, 1, 46) but : 100= 2*25+ 1*10+ 40
106=( 2, 2, 36) but : 100= 2*25+ 2*10+ 30
106=( 2, 3, 26) but : 100= 2*25+ 3*10+ 20
106=( 2, 4, 16) but : 100= 2*25+ 4*10+ 10
106=( 2, 5, 6) but : 100= 2*25+ 5*10+ 0
106=( 3, 0, 31) but : 100= 3*25+ 0*10+ 25
106=( 3, 1, 21) but : 100= 3*25+ 1*10+ 15
106=( 3, 2, 11) but : 100= 3*25+ 2*10+ 5
106=( 3, 3, 1) is good
106=( 4, 0, 6) but : 100= 4*25+ 0*10+ 0
The key features of the above given functions are correctly matched to their corresponding definition.
<h3>Definition of terms</h3>
- Negative sections of the graph: They are the parts where the graph is below the x-axis. That is option C.
- End behaviour: This is what happens to the graph on the far left or far right. That is option E.
- Positive sections of the graph: This is the parts where the graph is above the x-axis. That is option D.
- Intercepts: This is the points where the graph crosses an axis. That is option B
- Relative extrema: This is the points of relative minimum or maximum in a graph. That is option A.
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Answer:
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Step-by-step explanation:
