Answer:
$49.60
Step-by-step explanation:
The equation setup I used for this was as follows: $16 x (112/36)
Using PEMDAS, you'd start with the division part, 112/36. Although, I did this a bit differently, and being a bit further in school, I don't know if you go through this where you're at.
Start by factoring 112 into 4(28). This leaves you with 4(28)/36.
Next, cancel 4 out of 36. Since 4 is the common factor in 28 and 36, you cancel the four out of the equation, leaving you with 28/9. Convert this to its decimal form, 3.1.
Finally, take the 3.1 and multiply that by $16, which comes out to $49.60.
Answer:
-0.9090... can be written as
.
Explanation:
Any <em>repeating </em>decimal can be written as a fraction by dividing the section of the pattern to be repeated <em>by </em>9's.
We can start by listing out
0.909090... = 9/10 + 0/100 + 9/1000 + 0/10000 + 9/100000 + 0/1000000 + ...
Now. we let this series be equal to x, that is
= 9/10 + 0/100 + 9/1000 + 0/10000 + 9/100000 + 0/1000000 + ...
Now, we'll multiply both sides by 100
.
= 90 + 0 + 9/10 + 0/100 + 9/1000 + 0/10000 + ...
Then, subtract the 1st equation from the second like so:
= 90 + 0 + 9/10 + 0/100 + 9/1000 + 0/10000 + 9/100000 + 0/1000000 + ...
= - 9/10 - 0/100 - 9/1000 - 0/10000 - 9/100000 - 0/1000000 - ...
And we end up with this:

Finally, we divide both sides by 99 in order to isolate x and get the fraction we're looking for.

Which can be reduced and simplified to

Hope this helps!
Answer:
2.4 pounds of blueberries for each pie
Step-by-step explanation:
Mary has 12 pounds of blueberries to make 5 pies.
Divide 12 by 5 to see how much pounds of blueberries each pie will have.
There are 2.4 pounds of blueberries for each pie.
Answer:
![\left[\begin{array}{cc}2&8\\5&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%268%5C%5C5%261%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
The <em>transpose of a matrix </em>
is one where you swap the column and row index for every entry of some original matrix
. 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
and
to refer to the entry in the i-th row and the j-th column of the matrices
and
respectively:

Constructing the matrix
from those entries gives us
![P^T=\left[\begin{array}{cc}2&8\\5&1\end{array}\right]](https://tex.z-dn.net/?f=P%5ET%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%268%5C%5C5%261%5Cend%7Barray%7D%5Cright%5D)
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!
Answer:
6 hours and 15 minutes
Step-by-step explanation:
4 hours + 45 minutes + 1 hour + 30 minutes = 5 hours + 75 minutes = 6 hours + 15 minutes