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sergey [27]
3 years ago
9

5/6 + 1/3 adding fractions with unlike denominators.

Mathematics
1 answer:
Phantasy [73]3 years ago
4 0

Answer: 7/6 or 1 1/6

Step-by-step explanation:

First, convert all the fractions so that they have like denominators, using LCM, or Least Common Multiple. This means that you must find the least common multiple that the two denominators share, which in this case is 6. 6 will become the denominator for both fractions  At this point, to find the numerators, I use my own sort of mental method to get the answer. I first take the LCM and divide it by the denominator of one of the fractions (let's do 1/3) which would give me 2. Then multiply this number by the numerator of the same fraction you started with, which is 2 again. So, your new fraction is 2/6. Since 5/6 already has a denominator of 6, you don't have to do anything, but if it did, you would just repeat the same method. At this point, the new problem is 5/6 + 2/6. Add like normal, and leave in either improper form (7/6) or mixed numbers (1 1/6) based on instructions.

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2 years ago
Write a life situation for the inequality x&lt;2
iVinArrow [24]

Answer:

see below (I hope this helps!)

Step-by-step explanation:

A real-life situation for this inequality could be "Tom runs a lemonade stand. His profit is x. If Tom knows that his profit is less than 2 dollars, what inequality represents this situation?"

3 0
3 years ago
Please help ASAP brainliest.
lisov135 [29]

Answer:

\left[\begin{array}{cc}2&8\\5&1\end{array}\right]

Step-by-step explanation:

The <em>transpose of a matrix </em>M^T is one where you swap the column and row index for every entry of some original matrix M. Let's go through our first matrix row by row and swap the indices to construct this new matrix. Note that entries with the same index for row and column will stay fixed. Here I'll use the notation p_{i,j} and p^T_{i,j} to refer to the entry in the i-th row and the j-th column of the matrices P and P^T respectively:

p_{1,1}=p^T_{1,1}=2\\p_{1,2}=p^T_{2,1}=5\\p_{2,1}=p^T_{1,2}=8\\p_{2,2}=p^T_{2,2}=1\\

Constructing the matrix P^T from those entries gives us

P^T=\left[\begin{array}{cc}2&8\\5&1\end{array}\right]

which is option a. from the list.

Another interesting quality of the transpose is that we can geometrically represent it as a reflection over the line traced out by all of the entries where the row and column index are equal. In this example, reflecting over the line traced from 2 to 1 gives us our transpose. For another example of this, see the attached image!

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