If it is #17 this:
the class sold 45 sandwiches and collected $150.75, all you need to do is divide.
150.75/45
= 3.35
$3.35 per sandwitch
sinx-✓(1-3sin2x)=0
-✓(1-3sin2x)=-sinx
apply squared both sides
(-✓(1-3sin2x)^2=(-sinx)^2
1-3sin2x=sin2x
collect like terms
-3sin2x-sin2x+1=0
-4sin2x+1=0
-4sin2x=-1
devide both sides by -4
sin2x=-1/-4
sin2x=0.25
sinx*sinx =0.25
[sinx]^2 = 0.25
apply square root both sides
<h2>✓(sinx)2 = ✓0.25</h2>
<h2>sinx=0.5</h2><h2> </h2><h2>x=sin^-(0.5)</h2>
<h2>x=30°</h2>
<h3>check quadrant where sin is positive, sin is +ve in second quandrant</h3>
180-x= Theta(X)
180-30=X
X=150°
therefore, all angles for sinx -✓(1-3sin2x)=0 are (X= 30° and 150°)
An example of an item where you would need to find the area of a square is a square table.
An example of an item where you would need to find the area of a rectangle is a phone that has a rectangular shape.
<h3>How to calculate the area?</h3>
It's important to note that the area of a square is the multiplication of its sides by itself. For example, if the side is 4cm, the area will be:
= 4²
= 4 × 4.
= 16cm²
The area of a rectangle will be:
= Length × Width
Assuming length and width are 5cm and 2cm. This will be:
= 5 × 2
= 10cm²
Learn more about area on:
brainly.com/question/25292087
#SPJ1
![\huge \boxed{\mathbb{QUESTION} \downarrow}](https://tex.z-dn.net/?f=%20%5Chuge%20%5Cboxed%7B%5Cmathbb%7BQUESTION%7D%20%5Cdownarrow%7D)
![\begin{bmatrix} \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \end{bmatrix} \begin{bmatrix} \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { - 1 } & { 1 } & { 5 } \end{array} \end{bmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bbmatrix%7D%20%5Cbegin%7Barray%7D%20%7B%20l%20l%20%7D%20%7B%202%20%7D%20%26%20%7B%203%20%7D%20%5C%5C%20%7B%205%20%7D%20%26%20%7B%204%20%7D%20%5Cend%7Barray%7D%20%5Cend%7Bbmatrix%7D%20%5Cbegin%7Bbmatrix%7D%20%5Cbegin%7Barray%7D%20%7B%20l%20l%20l%20%7D%20%7B%202%20%7D%20%26%20%7B%200%20%7D%20%26%20%7B%203%20%7D%20%5C%5C%20%7B%20-%201%20%7D%20%26%20%7B%201%20%7D%20%26%20%7B%205%20%7D%20%5Cend%7Barray%7D%20%5Cend%7Bbmatrix%7D)
![\large \boxed{\mathbb{ANSWER\: WITH\: EXPLANATION} \downarrow}](https://tex.z-dn.net/?f=%20%5Clarge%20%5Cboxed%7B%5Cmathbb%7BANSWER%5C%3A%20WITH%5C%3A%20EXPLANATION%7D%20%5Cdownarrow%7D)
![\begin{bmatrix} \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \end{bmatrix} \begin{bmatrix} \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { - 1 } & { 1 } & { 5 } \end{array} \end{bmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bbmatrix%7D%20%5Cbegin%7Barray%7D%20%7B%20l%20l%20%7D%20%7B%202%20%7D%20%26%20%7B%203%20%7D%20%5C%5C%20%7B%205%20%7D%20%26%20%7B%204%20%7D%20%5Cend%7Barray%7D%20%5Cend%7Bbmatrix%7D%20%5Cbegin%7Bbmatrix%7D%20%5Cbegin%7Barray%7D%20%7B%20l%20l%20l%20%7D%20%7B%202%20%7D%20%26%20%7B%200%20%7D%20%26%20%7B%203%20%7D%20%5C%5C%20%7B%20-%201%20%7D%20%26%20%7B%201%20%7D%20%26%20%7B%205%20%7D%20%5Cend%7Barray%7D%20%5Cend%7Bbmatrix%7D)
In matrix multiplication, the number of columns in the 1st matrix is equal to the number of rows in the 2nd matrix.
![\left(\begin{matrix}2&3\\5&4\end{matrix}\right)\left(\begin{matrix}2&0&3\\-1&1&5\end{matrix}\right)](https://tex.z-dn.net/?f=%5Cleft%28%5Cbegin%7Bmatrix%7D2%263%5C%5C5%264%5Cend%7Bmatrix%7D%5Cright%29%5Cleft%28%5Cbegin%7Bmatrix%7D2%260%263%5C%5C-1%261%265%5Cend%7Bmatrix%7D%5Cright%29%20)
Multiply each element of the 1st row of the 1st matrix by the corresponding element of the 1st column of the 2nd matrix. Then add these products to obtain the element in the 1st row, 1st column of the product matrix.
![\left(\begin{matrix}2\times 2+3\left(-1\right)&&\\&&\end{matrix}\right)](https://tex.z-dn.net/?f=%5Cleft%28%5Cbegin%7Bmatrix%7D2%5Ctimes%202%2B3%5Cleft%28-1%5Cright%29%26%26%5C%5C%26%26%5Cend%7Bmatrix%7D%5Cright%29%20)
The remaining elements of the product matrix are found in the same way.
![\left(\begin{matrix}2\times 2+3\left(-1\right)&3&2\times 3+3\times 5\\5\times 2+4\left(-1\right)&4&5\times 3+4\times 5\end{matrix}\right)](https://tex.z-dn.net/?f=%5Cleft%28%5Cbegin%7Bmatrix%7D2%5Ctimes%202%2B3%5Cleft%28-1%5Cright%29%263%262%5Ctimes%203%2B3%5Ctimes%205%5C%5C5%5Ctimes%202%2B4%5Cleft%28-1%5Cright%29%264%265%5Ctimes%203%2B4%5Ctimes%205%5Cend%7Bmatrix%7D%5Cright%29%20)
Simplify each element by multiplying the individual terms.
![\left(\begin{matrix}4-3&3&6+15\\10-4&4&15+20\end{matrix}\right)](https://tex.z-dn.net/?f=%5Cleft%28%5Cbegin%7Bmatrix%7D4-3%263%266%2B15%5C%5C10-4%264%2615%2B20%5Cend%7Bmatrix%7D%5Cright%29%20)
Now, sum each element of the matrix.
![\large\boxed{\boxed{\left(\begin{matrix}1&3&21\\6&4&35\end{matrix}\right) }}](https://tex.z-dn.net/?f=%20%5Clarge%5Cboxed%7B%5Cboxed%7B%5Cleft%28%5Cbegin%7Bmatrix%7D1%263%2621%5C%5C6%264%2635%5Cend%7Bmatrix%7D%5Cright%29%20%7D%7D)