now, answer this question: given $ \triangle j k l $ with vertices at $ j ( - 2, - 1 ) $, $ k ( 6,0 ) $, and $ l ( - 2,5 ) $, prove that $ \triangle j k l $ is scalene. round your answer to the nearest hundredths. use the keypad to enter the answers in the boxes provided. the length of $ j k $ is $ \square $ units. the length of $ k l $ is $ \square $ units. the length of $ j l $ is $ \square $ units. since no two sides are equal, we proved that $ \triangle j k l $ is scalene.

Question

Accepted Answer

# Explanation:
## Step1: Calculate the length of $JK$
Use the distance formula $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. For $J(-2,-1)$ and $K(6,0)$, $x_1=-2,y_1 = - 1,x_2=6,y_2 = 0$.
$$
\begin{align*}
JK&=\sqrt{(6-(-2))^2+(0 - (-1))^2}\
&=\sqrt{(6 + 2)^2+(0+1)^2}\
&=\sqrt{8^2+1^2}\
&=\sqrt{64 + 1}\
&=\sqrt{65}\approx8.06
\end{align*}
$$
## Step2: Calculate the length of $KL$
For $K(6,0)$ and $L(-2,5)$, $x_1 = 6,y_1=0,x_2=-2,y_2 = 5$
$$
\begin{align*}
KL&=\sqrt{(-2 - 6)^2+(5-0)^2}\
&=\sqrt{(-8)^2+5^2}\
&=\sqrt{64+25}\
&=\sqrt{89}\approx9.43
\end{align*}
$$
## Step3: Calculate the length of $JL$
For $J(-2,-1)$ and $L(-2,5)$, $x_1=-2,y_1=-1,x_2=-2,y_2 = 5$
$$
\begin{align*}
JL&=\sqrt{(-2-(-2))^2+(5-(-1))^2}\
&=\sqrt{(-2 + 2)^2+(5 + 1)^2}\
&=\sqrt{0^2+6^2}\
&=6
\end{align*}
$$

# Answer:
The length of $JK$ is $8.06$ units.
The length of $KL$ is $9.43$ units.
The length of $JL$ is $6.00$ units.

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QUESTION IMAGE

Question

Explanation:

Step1: Calculate the length of \(JK\)

Step2: Calculate the length of \(KL\)

Step3: Calculate the length of \(JL\)

Answer: