Is there enough information to conclude that the two triangles are congruent? If so, what is a correct congruence statement?
2 answers:
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Answer:
Step-by-step explanation:
To find the midpoint, you must average the x's then average the y's.
So the x's are and
.
The y's are and
.
To find an average of a data with 2 elements you must add the two elements then take that sum and divide it by 2.
So this is what we will do with the x's then the y's.
So first x:
Write as improper fractions:
Division by 2 is the same as multiplying by 1/2:
The least common multiple of 6 and 10 is 30. We will multiply first fraction by 5/5 and 3/3 for the second so we can have the same denominator.
Now that the bottoms are the same we can write as one fraction:
Simplify:
If you want the answer as a mix fraction like your question began as we can do that by first figuring out how many 30's are in 32 and what is the remainder of that division.
There is one 30 and 32 and so 32-30=2 is the remainder.
Reduce by dividing top and bottom by 2:
Now for y:
Write the mix fractions as improper fractions:
Dividing by 2 is the same as multiplying by 1/2:
The least common multiple of 12 and 6 is 12 so I'm going to multiply the second fraction by 2/2 so the denominators will be the same:
Since the denominators are the same we can write as a single fraction:
How many 12's are in 41? 3 and so the remainder is 41-3(12)=41-36=5.
So the midpoint is:
.
