$\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ \sqrt{7}}}\quad ,&{{ -\sqrt{2}}})\quad 
% (c,d)
&({{4\sqrt{7}}}\quad ,&{{ 5\sqrt{2}}})
\end{array}~~~
% distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
$

$\bf d=\sqrt{[4\sqrt{7}-\sqrt{7}]^2~+~[5\sqrt{2}-(-\sqrt{2})]^2}
\\\\\\
d=\sqrt{(4\sqrt{7}-\sqrt{7})^2~+~(5\sqrt{2}+\sqrt{2})^2}\implies d=\sqrt{(3\sqrt{7})^2~+~(6\sqrt{2})^2}
\\\\\\
d=\sqrt{3^2\cdot 7~~+~~6^2\cdot 2}\implies d=\sqrt{63+72}\implies d=\sqrt{135}
\\\\\\
\begin{cases}
135=5\cdot 3\cdot 3\cdot 3\\
\qquad 5\cdot 3^2\cdot 3\\
\qquad 15\cdot 3^2
\end{cases}\implies d=\sqrt{15\cdot 3^2}\implies d=3\sqrt{15}$

8 0

3 years ago

A traffic helicopter descends 153 meters to be 445 meters above the ground, as illustrated in the

sertanlavr [38]

Answer:

598 meters

Step-by-step explanation:

In the problem, the word 'decend' means to decrease in elevation.

Our equation:

$? - 153 = 445$

Add 153 to both sides of the equation:

$?= 598$

598 meters represents the original height of the helicopter.

6 0

2 years ago