How do you divide mixed fractions
2 answers:
To divide fractions, we can use a method called "copy dot flip."
Let's say we were given 1/4 ÷ 1/2.
Copy: the first fraction.
Dot: change from dividing to multiplying.
Flip: take the reciprocal of the last fraction.
________
Copy: 1/4
Dot: · (×)
Flip: 2/1
Now, write as an equation.
1/4 · 2/1
________
Now, to multiply fractions, you multiply the number right next to it.
Simplify.
2/4 or 1/2
________
Best Regards,
Wolfyy :)
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Yes, we can obtain a diagonal matrix by multiplying two non diagonal matrix.
Consider the matrix multiplication below
![\left[\begin{array}{cc}a&b\\c&d\end{array}\right] \left[\begin{array}{cc}e&f\\g&h\end{array}\right] = \left[\begin{array}{cc}a e+b g&a f+b h\\c e+d g&c f+d h\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7De%26f%5C%5Cg%26h%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%20e%2Bb%20g%26a%20f%2Bb%20h%5C%5Cc%20e%2Bd%20g%26c%20f%2Bd%20h%5Cend%7Barray%7D%5Cright%5D%20)
For the product to be a diagonal matrix,
a f + b h = 0 ⇒ a f = -b h
and c e + d g = 0 ⇒ c e = -d g
Consider the following sets of values
The the matrix product becomes:
![\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}\frac{1}{3}&-1\\-\frac{1}{4}&\frac{1}{2}\end{array}\right] = \left[\begin{array}{cc}\frac{1}{3}-\frac{1}{2}&-1+1\\1-1&-3+2\end{array}\right]= \left[\begin{array}{cc}-\frac{1}{6}&0\\0&-1\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%262%5C%5C3%264%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Cfrac%7B1%7D%7B3%7D%26-1%5C%5C-%5Cfrac%7B1%7D%7B4%7D%26%5Cfrac%7B1%7D%7B2%7D%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Cfrac%7B1%7D%7B3%7D-%5Cfrac%7B1%7D%7B2%7D%26-1%2B1%5C%5C1-1%26-3%2B2%5Cend%7Barray%7D%5Cright%5D%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-%5Cfrac%7B1%7D%7B6%7D%260%5C%5C0%26-1%5Cend%7Barray%7D%5Cright%5D)
Thus, as can be seen we can obtain a diagonal matrix that is a product of non diagonal matrices.
Consider the matrix multiplication below
For the product to be a diagonal matrix,
a f + b h = 0 ⇒ a f = -b h
and c e + d g = 0 ⇒ c e = -d g
Consider the following sets of values
The the matrix product becomes:
Thus, as can be seen we can obtain a diagonal matrix that is a product of non diagonal matrices.
Part A
The notation means that we're approaching x = 2 from the right hand side (aka positive side). This is known as a right hand limit.
So we could start at say x = 2.5 and get closer to 2 by getting to x = 2.4 then to x = 2.3 then 2.2, 2.1, 2.01, 2.001, etc
We don't actually arrive at x = 2 itself. We simply move closer and closer.
Since we're on the positive or right hand side of 2, this means we go with the rule involving x > 2
Therefore f(x) = (x/2) + 1
Plug in x = 2 to find that...
f(x) = (x/2) + 1
f(2) = (2/2) + 1
f(2) = 2
This shows
Then for the left hand limit , we'll involve x < 2 and we go for the first piece. So,
f(x) = 3-x
f(2) = 3-2
f(2) = 1
Therefore,
===============================================================
Part B
Because this means that the limit
does not exist.
If you are a visual learner, check out the graph below of the piecewise function. Notice the gap or disconnect at x = 2. This can be thought of as two roads that are disconnected. There's no way for a car to go from one road to the other. Because of this disconnect, the limit doesn't exist at x = 2.
===============================================================
Part C
You'll follow the same type of steps shown in part A.
However, keep in mind that x = 4 is above x = 2, so we'll deal with x > 2 only.
So you'd only involve the second piece f(x) = (x/2) + 1
You should find that f(4) = 3, and that both left and right hand limits equal this value. The left and right hand limits approach the same y value. The limit does exist here. There are no gaps to worry about when x = 4.
===============================================================
Part D
As mentioned earlier, since , this means the limit
does exist and it's equal to 3.
As x gets closer and closer to 4, the y values are approaching 3. This applies to both directions.
