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:
3/12: terminating; 2/9: repeating
Step-by-step explanation:
3/12=0.25 terminating as it has a finite number of digits after the decimal point.
2/9=0.222.... repeating decimal as its number is repeated indefinitely.
Answer:
y = ± 
Step-by-step explanation:
Given
x = a - 2by² ( add 2by² to both sides )
x + 2by² = a ( subtract x from both sides )
2by² = a - x ( divide both sides by 2b )
y² =
( take the square root of both sides )
y = ± 
Step-by-step explanation:
P(X)=2x²-5x-3 is in the form ax²+bx+c
Using quadratic equation
x={-b±√(b²-4ac)}/2a
x=3,-1/2
The missing X value equals 5. The triangle was dilated by a scale factor of 3 and 15 / 3 = 5.