Cos B = c/a
cosine is adjacent over hypotenuse.
there is an acronym for this: soh-cah-toa
soh: sine is opposite over hypotenuse
cah: cosine is adjacent over hypotenuse
toa: tangent is opposite over adjacent
Answer:
r=3
Step-by-step explanation:
Simplify the expression
16-2r=3r+1
isolate r by adding 2r to both sides
16=5r+1
isolate the numbers by subtracting 1 from both sides
15=5r
divide both sides by 5 to completely isolate r
r=3
Answer:
The products of AB and BA is given by
AB=![\left[\begin{array}{cc}21&-6\\ 9&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D21%26-6%5C%5C%209%263%5Cend%7Barray%7D%5Cright%5D)
BA=![\left[\begin{array}{cc}9&-1\\-18&15\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D9%26-1%5C%5C-18%2615%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
Given the matrices A=
and
B=![\left[\begin{array}{cc}1&-1\\-3&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26-1%5C%5C-3%260%5Cend%7Barray%7D%5Cright%5D)
To find the product AB and BA
AB=![\left[\begin{array}{cc}6&-5\\-3&-4\end{array}\right] \left[\begin{array}{cc}1 & -1\\-3 &0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%26-5%5C%5C-3%26-4%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%20%26%20-1%5C%5C-3%20%260%5Cend%7Barray%7D%5Cright%5D)
![=\left[\begin{array}{cc}6+15& -6+0\\-3+12&3-0\end{array}\right]](https://tex.z-dn.net/?f=%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%2B15%26%20-6%2B0%5C%5C-3%2B12%263-0%5Cend%7Barray%7D%5Cright%5D)
Therefore the product of AB is
AB=![\left[\begin{array}{cc}21&-6\\ 9&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D21%26-6%5C%5C%209%263%5Cend%7Barray%7D%5Cright%5D)
BA=
![\left[\begin{array}{cc}6&-5\\-3&-4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%26-5%5C%5C-3%26-4%5Cend%7Barray%7D%5Cright%5D)
![=\left[\begin{array}{cc}6+3& -5+4\\-18+0&15+0\end{array}\right]](https://tex.z-dn.net/?f=%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%2B3%26%20-5%2B4%5C%5C-18%2B0%2615%2B0%5Cend%7Barray%7D%5Cright%5D)
Therefore the product of BA is
BA=![\left[\begin{array}{cc}9&-1\\-18&15\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D9%26-1%5C%5C-18%2615%5Cend%7Barray%7D%5Cright%5D)
The products of AB and BA is given by
AB=![\left[\begin{array}{cc}21&-6\\ 9&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D21%26-6%5C%5C%209%263%5Cend%7Barray%7D%5Cright%5D)
BA=![\left[\begin{array}{cc}9&-1\\-18&15\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D9%26-1%5C%5C-18%2615%5Cend%7Barray%7D%5Cright%5D)
Answer: x = 9, y = -14
Explanation:
4x + y = 16
2x + y = -2<span>
</span> 4x + y = 16
- 2x + y = -2
-------------------
2x + 0 = 18
-------------------
2x = 18
x = 9 <<<<< ¯\_(ツ)_/¯ For the value of x
Plug x = 9 into 2x + y = -2
2(9) + y = -2
12 + y = -2
y = -14 <<<<< ¯\_(ツ)_/¯ For the value of y