The distance between two points on the plane is given by the formula below
![\begin{gathered} A=(x_1,y_1),B=(x_2,y_2) \\ \Rightarrow d(A,B)=\sqrt[]{(x_1-x_2)^2+(y_1-y_2)^2} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20A%3D%28x_1%2Cy_1%29%2CB%3D%28x_2%2Cy_2%29%20%5C%5C%20%5CRightarrow%20d%28A%2CB%29%3D%5Csqrt%5B%5D%7B%28x_1-x_2%29%5E2%2B%28y_1-y_2%29%5E2%7D%20%5Cend%7Bgathered%7D)
Therefore, in our case,

Thus,
![\begin{gathered} \Rightarrow d(A,B)=\sqrt[]{(-1-5)^2+(-3-2)^2}=\sqrt[]{6^2+5^2}=\sqrt[]{36+25}=\sqrt[]{61} \\ \Rightarrow d(A,B)=\sqrt[]{61} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5CRightarrow%20d%28A%2CB%29%3D%5Csqrt%5B%5D%7B%28-1-5%29%5E2%2B%28-3-2%29%5E2%7D%3D%5Csqrt%5B%5D%7B6%5E2%2B5%5E2%7D%3D%5Csqrt%5B%5D%7B36%2B25%7D%3D%5Csqrt%5B%5D%7B61%7D%20%5C%5C%20%5CRightarrow%20d%28A%2CB%29%3D%5Csqrt%5B%5D%7B61%7D%20%5Cend%7Bgathered%7D)
Therefore, the answer is sqrt(61)
In general,

Remember that

Therefore,
Answer:
x > 22
Step-by-step explanation:
<em>Hey there!</em>
Well to solve,
52 - 3x < -14
we need to single out x
52 - 3x < -14
-52 to both sides
-3x < -66
Divide both sides by -3
x > 22
The < changes to > because the variable number is a - being divided.
<em>Hope this helps :)</em>
Answer:
I think it is 3.1
Step-by-step explanation:
make me brainlyest pls
Answer: 500
Step-by-step explanation:
The reason why it's 500 because it's already been rounded. 500 stays as 500 bc numbers 4 and below goes down to 0 when rounding. When rounding 0-4 with just the ones place it stays the same.
X = 5
Move the variable then cells to like terms and simplify