B. Five cubed because there at 5 2s there and 2*2*2*2*2 would be the same thing as saying 5^2
Answer:
first one: yes
second one: yes
third one: no
fourth one: yes
Step-by-step explanation:
Treat the matrices on the right side of each equation like you would a constant.
Let 2<em>X</em> + <em>Y</em> = <em>A</em> and 3<em>X</em> - 4<em>Y</em> = <em>B</em>.
Then you can eliminate <em>Y</em> by taking the sum
4<em>A</em> + <em>B</em> = 4 (2<em>X</em> + <em>Y</em>) + (3<em>X</em> - 4<em>Y</em>) = 11<em>X</em>
==> <em>X</em> = (4<em>A</em> + <em>B</em>)/11
Similarly, you can eliminate <em>X</em> by using
-3<em>A</em> + 2<em>B</em> = -3 (2<em>X</em> + <em>Y</em>) + 2 (3<em>X</em> - 4<em>Y</em>) = -11<em>Y</em>
==> <em>Y</em> = (3<em>A</em> - 2<em>B</em>)/11
It follows that
![X=\dfrac4{11}\begin{bmatrix}12&-3\\10&22\end{bmatrix}+\dfrac1{11}\begin{bmatrix}7&-10\\-7&11\end{bmatrix} \\\\ X=\dfrac1{11}\left(4\begin{bmatrix}12&-3\\10&22\end{bmatrix}+\begin{bmatrix}7&-10\\-7&11\end{bmatrix}\right) \\\\ X=\dfrac1{11}\left(\begin{bmatrix}48&-12\\40&88\end{bmatrix}+\begin{bmatrix}7&-10\\-7&11\end{bmatrix}\right) \\\\ X=\dfrac1{11}\begin{bmatrix}55&-22\\33&99\end{bmatrix} \\\\ X=\begin{bmatrix}5&-2\\3&9\end{bmatrix}](https://tex.z-dn.net/?f=X%3D%5Cdfrac4%7B11%7D%5Cbegin%7Bbmatrix%7D12%26-3%5C%5C10%2622%5Cend%7Bbmatrix%7D%2B%5Cdfrac1%7B11%7D%5Cbegin%7Bbmatrix%7D7%26-10%5C%5C-7%2611%5Cend%7Bbmatrix%7D%20%5C%5C%5C%5C%20X%3D%5Cdfrac1%7B11%7D%5Cleft%284%5Cbegin%7Bbmatrix%7D12%26-3%5C%5C10%2622%5Cend%7Bbmatrix%7D%2B%5Cbegin%7Bbmatrix%7D7%26-10%5C%5C-7%2611%5Cend%7Bbmatrix%7D%5Cright%29%20%5C%5C%5C%5C%20X%3D%5Cdfrac1%7B11%7D%5Cleft%28%5Cbegin%7Bbmatrix%7D48%26-12%5C%5C40%2688%5Cend%7Bbmatrix%7D%2B%5Cbegin%7Bbmatrix%7D7%26-10%5C%5C-7%2611%5Cend%7Bbmatrix%7D%5Cright%29%20%5C%5C%5C%5C%20X%3D%5Cdfrac1%7B11%7D%5Cbegin%7Bbmatrix%7D55%26-22%5C%5C33%2699%5Cend%7Bbmatrix%7D%20%5C%5C%5C%5C%20X%3D%5Cbegin%7Bbmatrix%7D5%26-2%5C%5C3%269%5Cend%7Bbmatrix%7D)
Similarly, you would find
![Y=\begin{bmatrix}2&1\\4&4\end{bmatrix}](https://tex.z-dn.net/?f=Y%3D%5Cbegin%7Bbmatrix%7D2%261%5C%5C4%264%5Cend%7Bbmatrix%7D)
You can solve the second system in the same fashion. You would end up with
![P=\begin{bmatrix}2&-3\\0&1\end{bmatrix} \text{ and } Q=\begin{bmatrix}1&2\\3&-1\end{bmatrix}](https://tex.z-dn.net/?f=P%3D%5Cbegin%7Bbmatrix%7D2%26-3%5C%5C0%261%5Cend%7Bbmatrix%7D%20%5Ctext%7B%20and%20%7D%20Q%3D%5Cbegin%7Bbmatrix%7D1%262%5C%5C3%26-1%5Cend%7Bbmatrix%7D)
Answer:
x=4
Step-by-step explanation:
Consider ΔKLJ:
sinθ=opposite/hypotenuse
sin30=LJ/8√2
LJ=0.5*8√2=4√2
Consider ΔMLJ:
sinθ=opp/hyp
sin45=x/4√2
1/√2=x/4√2
4√2/√2=x
x=4
Divide 327 by 12 to get 27.25. Howard pays 27 dollars and 25 cents for each card.