You left out the 9 in the whole equation. The 1st step you had right you just forgot to put +9. 2nd step you are combining the like terms and get 11-26x=-34x+40. Step 3: you are trying to isolate the x on one of the sides. You need to switch the 11 over to the 40 and get 29. Move -34x over to -26x and get 8x. Step 4 all you are doing in dividing 8 from both sides and get 5.
Answer:
Step-by-step explanation:
For the orange table:
<em> </em>Substitute the corresponding values of <em>x</em> into the function 
, then:
For a:
x=0
For b:
x=2
For c:
x=4
For the blue table:
<em> </em>Substitute the corresponding values of <em>x</em> into the function 
, then:
For d:
x=0
For e:
x=2
For f:
x=4
__ 5 __, __
the first would be 2
after 5 is 8
after 8 is 11
Hope this helps!!!
Answer:
700.4 cm
Step-by-step explanation:
This involves two similar triangles.
Both triangles are right triangles.
One has legs measuring 1 cm and 30 cm. We can find the hypotenuse by using the Pythagorean theorem.
(1 cm)^2 + (30 cm)^2 = c^2
c^2 = 901 cm^2
c = sqrt(901) cm
The second triangle has one leg with length 700 cm. This leg corresponds to the 30-cm leg in the other triangle. Since the triangles are similar, we can use a proportion to find the hypotenuse of the second triangle.
(30 cm)/(700 cm) = [sqrt(901) cm]/x
3/70 = sqrt(901) cm/x
3x = 70 * sqrt(901) cm
x = 70 * sqrt(901) cm/3
x = 700.4 cm
Answer: 700.4 cm
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!
