$\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
A(\stackrel{x_1}{4}~,~\stackrel{y_1}{5})\qquad 
B(\stackrel{x_2}{1}~,~\stackrel{y_2}{2})\qquad \qquad 
% distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
AB=\sqrt{(1-4)^2+(2-5)^2}\implies AB=\sqrt{(-3)^2+(-3)^2}
\\\\\\
AB=\sqrt{9+9}\implies \boxed{AB=\sqrt{18}}\implies AB=\sqrt{2(9)}
\\\\\\
AB=\sqrt{2(3^2)}\implies \implies AB=3\sqrt{2}$

5 0

3 years ago

Solve this : <br> 1. <img src="https://tex.z-dn.net/?f=6" id="TexFormula1" title="6" alt="6" align="absmiddle" class="latex-form

leonid [27]

6/2 (2+1)

6/ 2 x (3)

6/6 = 1

5 0

4 years ago

If B=3x^{2}-x+3 and A=x-6, find an expression that equals 2B+3A in standard form.

Ivan

9514 1404 393

Answer:

6x^2 +x -12

Step-by-step explanation:

Substitute for A and B and collect terms.

2B +3A

= 2(3x^2 -x +3) +3(x -6) . . . . substitute for A and B

= 6x^2 -2x +6 +3x -18 . . . . . eliminate parentheses

= 6x^2 +x -12 . . . . . . . . . . . . collect terms

7 0

3 years ago