(2/3),=(1/2), 3,4
.
............................
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,
The square root of infinity is still infinity, or rather, undefined or indeterminate. For example, if you take the square root of a big number, you will
still have a big number as a result, now if you make the number even
bigger, the square root will also get bigger, now imagine what will
happen if this number is infinite.
F(x) =x² - 10x, f⁻¹(x) =?
1st find the missing square of x²-10x, ==> (x-5)² - 25
y= (x-5)² - 25; replace x by y and v0ce versa: x= (y-5)² -25
or
(y-5)² = x+25
y-5 = √(x+25) and y = √(x+25) -5
Domain = {x∈R: X>= 25} AND Range ={y∈R: y>= 5}
Since only one person can finish in first place, one in second, etc., the total number of placers is 12x11x10, or 1320 different ways the 12 racers can finish 1st, 2nd and 3rd....
☺☺☺☺