Which is greater .513 or .5129
2 answers:
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Let's compare the given function with the model for a quadratic equation:
Since the value of a is positive, the parabola has its concavity upwards, and the function has a minimum value.
The minimum value can be found calculating the y-coordinate of the vertex:
Therefore the minimum value is -24.
Answer:
The claim is correct
Explanation:
Assume the given triangle ABC
perimeter of triangle ABC = AB + BC + CA ............> I
Now, we have:
D is the midpoint of AB, this means that:
AD = DB = (1/2) AB ..........> 1
E is the midpoint of AC, this means that:
AE = EC = (1/2) AC ...........> 2
DE is the midsegment in triangle ABC, this means that:
DE = (1/2) BC ...........> 3
perimeter of triangle ADE = AD + DE + EA
Substitute in this equation with the corresponding lengths in 1,2 and 3:
perimeter of triangle ADE = (1/2) AB + (1/2) BC = (1/2) AC
perimeter of triangle ADE = (1/2)(AB+BC+AC) .........> II
From I and II, we can prove that:
perimeter of triangle ADE = (1/2) perimeter of triangle ABC
Which means that:
perimeter of midsegment triangle is half the perimeter of the original triangle.
Hope this helps :)
The claim is correct
Explanation:
Assume the given triangle ABC
perimeter of triangle ABC = AB + BC + CA ............> I
Now, we have:
D is the midpoint of AB, this means that:
AD = DB = (1/2) AB ..........> 1
E is the midpoint of AC, this means that:
AE = EC = (1/2) AC ...........> 2
DE is the midsegment in triangle ABC, this means that:
DE = (1/2) BC ...........> 3
perimeter of triangle ADE = AD + DE + EA
Substitute in this equation with the corresponding lengths in 1,2 and 3:
perimeter of triangle ADE = (1/2) AB + (1/2) BC = (1/2) AC
perimeter of triangle ADE = (1/2)(AB+BC+AC) .........> II
From I and II, we can prove that:
perimeter of triangle ADE = (1/2) perimeter of triangle ABC
Which means that:
perimeter of midsegment triangle is half the perimeter of the original triangle.
Hope this helps :)