Answer:
Step-by-step explanation:
The distance between B and B' will be the same as distance between A and A'.
= sqrt ( (5-1)^2 + ( 1 - -2)^2)
= sqrt (16 + 9)
= 5 units answer
Answer:
-3
Step-by-step explanation:
f(x)= x²+3x-1
f(-1)= (-1)²+3(-1)-1
= 1-3-1
= -2-1
= -3
Answer:
lol
Step-by-step explanation:
hmmm this is messy
<h2>
Answer:</h2>
![\boxed{x=1, \y=-1, \ z=2}](https://tex.z-dn.net/?f=%5Cboxed%7Bx%3D1%2C%20%5Cy%3D-1%2C%20%5C%20z%3D2%7D)
<h2>
Step-by-step explanation:</h2>
We will use the Gaussian elimination method to solve this problem. To do so, let's follow the following steps:
Step 1: Let's multiply first equation by −2. Next, add the result to the second equation. So:
![\begin{array}{ cccc }~~ x&+~~2~ y&-~~~~~ z&~=~-3\\&-~~~5~ y&+~~3~ z&~=~11\\~~ x&-~~~~~ y&+~~~~ z&~=~4\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7B%20cccc%20%7D~~%20x%26%2B~~2~%20y%26-~~~~~%20z%26~%3D~-3%5C%5C%26-~~~5~%20y%26%2B~~3~%20z%26~%3D~11%5C%5C~~%20x%26-~~~~~%20y%26%2B~~~~%20z%26~%3D~4%5Cend%7Barray%7D)
Step 2: Let's multiply first equation by −1. Next, add the result to the third equation. Thus:
![\begin{array}{ cccc }~~ x&+~~2~ y&-~~~~~ z&~=~-3\\&-~~~5~ y&+~~3~ z&~=~11\\&-~~~3~ y&+~~2~ z&~=~7\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7B%20cccc%20%7D~~%20x%26%2B~~2~%20y%26-~~~~~%20z%26~%3D~-3%5C%5C%26-~~~5~%20y%26%2B~~3~%20z%26~%3D~11%5C%5C%26-~~~3~%20y%26%2B~~2~%20z%26~%3D~7%5Cend%7Barray%7D)
Step 3: Let's multiply second equation by −35, Next, add the result to the third equation. Therefore:
![\begin{array}{ cccc }~~ x&+~~2~ y&-~~~~~ z&~=~-3\\&-~~~5~ y&+~~3~ z&~=~11\\&&+~~\frac{ 1 }{ 5 }~ z&~=~\frac{ 2 }{ 5 }\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7B%20cccc%20%7D~~%20x%26%2B~~2~%20y%26-~~~~~%20z%26~%3D~-3%5C%5C%26-~~~5~%20y%26%2B~~3~%20z%26~%3D~11%5C%5C%26%26%2B~~%5Cfrac%7B%201%20%7D%7B%205%20%7D~%20z%26~%3D~%5Cfrac%7B%202%20%7D%7B%205%20%7D%5Cend%7Barray%7D)
Step 4: solve for z, then for y, then for x:
![\frac{ 1 }{ 5 } ~ z & = \frac{ 2 }{ 5 } \\ \\ \boxed{z & = 2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%201%20%7D%7B%205%20%7D%20~%20z%20%26%20%3D%20%5Cfrac%7B%202%20%7D%7B%205%20%7D%20%5C%5C%20%5C%5C%20%5Cboxed%7Bz%20%26%20%3D%202%7D)
![-5y+3z &= 11\\-5y+3\cdot 2 &= 11\\ \\ \boxed{y &= -1}](https://tex.z-dn.net/?f=-5y%2B3z%20%26%3D%2011%5C%5C-5y%2B3%5Ccdot%202%20%26%3D%2011%5C%5C%20%5C%5C%20%5Cboxed%7By%20%26%3D%20-1%7D)
By substituting
into the first equation, we get the
. So:
![x+2(-1)-2=-3 \\ \\ x-2-2=-3 \\ \\ \boxed{x=1}](https://tex.z-dn.net/?f=x%2B2%28-1%29-2%3D-3%20%5C%5C%20%5C%5C%20x-2-2%3D-3%20%5C%5C%20%5C%5C%20%5Cboxed%7Bx%3D1%7D)