The Given Triangle PMO is a Right Angled Triangle with m∠M = 90°
Given m∠P = 40°
We know that : Sum of Angles in a Triangle = 180°
⇒ m∠P + m∠M + m∠O = 180°
⇒ 40° + 90° + m∠O = 180°
⇒ 130° + m∠O = 180°
⇒ m∠O = 180° - 130°
⇒ m∠O = 50°
We can notice that m∠O and m∠1 form a Linear Pair (180°)
⇒ m∠O + m∠1 = 180°
⇒ 50° + m∠1 = 180°
⇒ m∠1 = 180° - 50°
⇒ m∠1 = 130°
Last Option is the Answer
Answer with explanation:
1. The given equations are
3x -5 y=2
-x+2 y= 0
⇒The matrix in the form of , AX=B, is
![A=\left[\begin{array}{cc}3&-5\\-1&2\end{array}\right] ,\\\\ X=\left[\begin{array}{c}x&y\end{array}\right],\\\\B=\left[\begin{array}{c}2&0\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%26-5%5C%5C-1%262%5Cend%7Barray%7D%5Cright%5D%20%2C%5C%5C%5C%5C%20X%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%26y%5Cend%7Barray%7D%5Cright%5D%2C%5C%5C%5C%5CB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%260%5Cend%7Barray%7D%5Cright%5D)
Adj.A=Transpose of cofactor of Matrix A
![Adj.A=\left[\begin{array}{cc}2&1\\5&3\end{array}\right] ,\\\\ |A|=6-5\\\\|A|=1\\\\\left[\begin{array}{c}x&y\end{array}\right]=\left[\begin{array}{cc}2&5\\1&3\end{array}\right] \times \left[\begin{array}{c}2&0\end{array}\right]\\\\x=4, y=2](https://tex.z-dn.net/?f=Adj.A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%261%5C%5C5%263%5Cend%7Barray%7D%5Cright%5D%20%2C%5C%5C%5C%5C%20%7CA%7C%3D6-5%5C%5C%5C%5C%7CA%7C%3D1%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%26y%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%265%5C%5C1%263%5Cend%7Barray%7D%5Cright%5D%20%5Ctimes%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%260%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5Cx%3D4%2C%20y%3D2)
2.
The given equations are
x+y-z=2
x+z=7
2 x +y+z=13
⇒The matrix in the form of , AX=B, is
![A=\left[\begin{array}{ccc}1&1&-1\\1&0&1\\2&1&1\end{array}\right]\\\\ X=\left[\begin{array}{ccc}x\\y\\z\end{array}\right]\\\\B= \left[\begin{array}{ccc}2\\7\\13\end{array}\right]\\\\\rightarrow X=A^{-1}B\\\\\rightarrow X=\frac{Adj.A}{|A|}\times B\\\\a_{11}=-1,a_{12}=1,a_{13}=1,a_{21}=-2,a_{22}=3,a_{23}=1,a_{31}=1,a_{32}=-2,a_{33}=-1\\\\|A|=1\times(0-1)-1\times(1-2)-1\times(1-0)\\\\=-1+1-1\\\\|A|=-1\\\\Adj.A=\left[\begin{array}{ccc}-1&-2&1\\1&3&-2\\1&1&-1\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%26-1%5C%5C1%260%261%5C%5C2%261%261%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%20X%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5CB%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%5C%5C7%5C%5C13%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5Crightarrow%20X%3DA%5E%7B-1%7DB%5C%5C%5C%5C%5Crightarrow%20X%3D%5Cfrac%7BAdj.A%7D%7B%7CA%7C%7D%5Ctimes%20B%5C%5C%5C%5Ca_%7B11%7D%3D-1%2Ca_%7B12%7D%3D1%2Ca_%7B13%7D%3D1%2Ca_%7B21%7D%3D-2%2Ca_%7B22%7D%3D3%2Ca_%7B23%7D%3D1%2Ca_%7B31%7D%3D1%2Ca_%7B32%7D%3D-2%2Ca_%7B33%7D%3D-1%5C%5C%5C%5C%7CA%7C%3D1%5Ctimes%280-1%29-1%5Ctimes%281-2%29-1%5Ctimes%281-0%29%5C%5C%5C%5C%3D-1%2B1-1%5C%5C%5C%5C%7CA%7C%3D-1%5C%5C%5C%5CAdj.A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%26-2%261%5C%5C1%263%26-2%5C%5C1%261%26-1%5Cend%7Barray%7D%5Cright%5D)
![\frac{Adj.A}{|A|}=\left[\begin{array}{ccc}1&2&-1\\-1&-3&2\\-1&-1&1\end{array}\right]\\\\X=A^{-1}B\\\\\left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}1&2&-1\\-1&-3&2\\-1&-1&1\end{array}\right]\times\left[\begin{array}{ccc}2\\7\\13\end{array}\right]\\\\x=3,y=3,z=4](https://tex.z-dn.net/?f=%5Cfrac%7BAdj.A%7D%7B%7CA%7C%7D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%26-1%5C%5C-1%26-3%262%5C%5C-1%26-1%261%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5CX%3DA%5E%7B-1%7DB%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%26-1%5C%5C-1%26-3%262%5C%5C-1%26-1%261%5Cend%7Barray%7D%5Cright%5D%5Ctimes%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%5C%5C7%5C%5C13%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5Cx%3D3%2Cy%3D3%2Cz%3D4)
Answer:
Step-by-step explanation:
5(y+4) = 4(y+5)
5y+20 = 4y + 20
5y = 4y
y=0
2(3x+16)
Step-by-step explanation:
Identify the expression equivalent to 4(x + x + 7) − 2x + 8 − 4
4(x + x + 7) − 2x + 8 − 4
When x=1 , the expression becomes 4(1+1+7)-2(1)+8-4=38
When x=2 , the expression becomes 4(2+2+7)-2(1)+8-4=44
Plug in x=1 and check with each expression
6x + 11 =6(1) +11= 16
3(x + 7) = 3(1+7)= 3(8)= 24
2(3x + 16) = 2(3(1)+16)= 38 , we got same answer when x=1, now check with x=2
2(3x + 16) = 2(3(2)+16)= 44
3x + 16= 3+16= 19
Answer: $208.95
Step-by-step explanation: $199 x .05= 9.95 + $199= $208.95
