(4) 64 tiles, (5)
, (6) ![N^{2}=S^{4}](https://tex.z-dn.net/?f=N%5E%7B2%7D%3DS%5E%7B4%7D)
Step-by-step explanation:
Given data: Side length of square tile = 1 ft
(4) Side length of square pool = 8 ft
![\text {Number of tiles} =\frac{\text { Area of square pool }}{\text { area of square tile }}](https://tex.z-dn.net/?f=%5Ctext%20%7BNumber%20of%20tiles%7D%20%3D%5Cfrac%7B%5Ctext%20%7B%20Area%20of%20square%20pool%20%7D%7D%7B%5Ctext%20%7B%20area%20of%20square%20tile%20%7D%7D)
![=\frac{8 \times 8}{1 \times 1}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B8%20%5Ctimes%208%7D%7B1%20%5Ctimes%201%7D)
= 64
Hence 64 tiles needed for the border.
(5) Suppose number of square tiles be N and side of square tub is s ft.
So, N = ![\frac{s \times s}{1 \times 1}](https://tex.z-dn.net/?f=%5Cfrac%7Bs%20%5Ctimes%20s%7D%7B1%20%5Ctimes%201%7D)
![\Rightarrow N=s^{2}](https://tex.z-dn.net/?f=%5CRightarrow%20N%3Ds%5E%7B2%7D)
So, the expression is
.
(6) The equivalent expression is
.
How To Solve It:
$13.55 x 0.04 = 0.54
$13.55 + 0.54 = $14.09
Answer:
$14.09
I hope this helps! :)
Answer:
(1) 68.811 L
Step-by-step explanation:
The amount being subtracted is ...
16 + 0.18 + 0.009 = 16.189 . . . . liters
The difference is ...
85.000 -16.189 = 68.111 . . . liters
__
<em>Additional comment</em>
cL is the SI unit "centiliters." The prefix "centi-" mean 1/100 or 0.01 units. Then 18 cL = 18 × 0.01 L = 0.18 L.
mL is the SI unit "milliliters." The prefix "milli-" means 1/1000 or 0.001 units. Then 9 mL = 9 × 0.001 L = 0.009 L.
Let's consider an arbitrary 2x2 matrix as an example,
![\mathbf A=\begin{bmatrix}\mathbf x&\mathbf y\end{bmatrix}=\begin{bmatrix}x_1&y_1\\x_2&y_2\end{bmatrix}](https://tex.z-dn.net/?f=%5Cmathbf%20A%3D%5Cbegin%7Bbmatrix%7D%5Cmathbf%20x%26%5Cmathbf%20y%5Cend%7Bbmatrix%7D%3D%5Cbegin%7Bbmatrix%7Dx_1%26y_1%5C%5Cx_2%26y_2%5Cend%7Bbmatrix%7D)
The columns of
![\mathbf A](https://tex.z-dn.net/?f=%5Cmathbf%20A)
are linearly independent if and only if the column vectors
![\mathbf x,\mathbf y](https://tex.z-dn.net/?f=%5Cmathbf%20x%2C%5Cmathbf%20y)
are linearly independent.
This is the case if the only way we can make a linear combination of
![\mathbf x,\mathbf y](https://tex.z-dn.net/?f=%5Cmathbf%20x%2C%5Cmathbf%20y)
reduce to the zero vector is to multiply the vectors by 0; that is,
![c_1\mathbf x+c_2\mathbf y=\mathbf 0](https://tex.z-dn.net/?f=c_1%5Cmathbf%20x%2Bc_2%5Cmathbf%20y%3D%5Cmathbf%200)
only by letting
![c_1=c_2=0](https://tex.z-dn.net/?f=c_1%3Dc_2%3D0)
.
A more concrete example: suppose
![\mathbf A=\begin{bmatrix}1&2\\4&8\end{bmatrix}](https://tex.z-dn.net/?f=%5Cmathbf%20A%3D%5Cbegin%7Bbmatrix%7D1%262%5C%5C4%268%5Cend%7Bbmatrix%7D)
Here,
![\mathbf x=\begin{bmatrix}1\\4\end{bmatrix}](https://tex.z-dn.net/?f=%5Cmathbf%20x%3D%5Cbegin%7Bbmatrix%7D1%5C%5C4%5Cend%7Bbmatrix%7D)
and
![\amthbf y=\begin{bmatrix}2\\8\end{bmatrix}](https://tex.z-dn.net/?f=%5Camthbf%20y%3D%5Cbegin%7Bbmatrix%7D2%5C%5C8%5Cend%7Bbmatrix%7D)
. Notice that we can get the zero vector by taking
![c_1=-2](https://tex.z-dn.net/?f=c_1%3D-2)
and
![c_2=1](https://tex.z-dn.net/?f=c_2%3D1)
:
![-2\begin{bmatrix}1\\4\end{bmatrix}+\begin{bmatrix}2\\8\end{bmatrix}=\begin{bmatrix}-2+2\\-8+8\end{bmatrix}=\mathbf 0](https://tex.z-dn.net/?f=-2%5Cbegin%7Bbmatrix%7D1%5C%5C4%5Cend%7Bbmatrix%7D%2B%5Cbegin%7Bbmatrix%7D2%5C%5C8%5Cend%7Bbmatrix%7D%3D%5Cbegin%7Bbmatrix%7D-2%2B2%5C%5C-8%2B8%5Cend%7Bbmatrix%7D%3D%5Cmathbf%200)
so the columns of
![\mathbf A](https://tex.z-dn.net/?f=%5Cmathbf%20A)
are not linearly independent, or linearly dependent.
Answer:
sorry I cant speak spanish
Step-by-step explanation: