Write in standard form Thirty- nine and fifteen hundredths
1 answer:
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Given :
A vector v = 10i - 24k.
To Find :
A vector of magnitude 3 in the direction of v = 10i - 24k.
Solution :
Unit vector in the direction of vector v is :
Now , vector of magnitude 3 in direction v is :
Hence , this is the required solution .
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 :)