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3 years ago

Evaluate 8-11 to the square root of 25/121

1 answer:

4 0

$\frac{8-11}{\sqrt{\frac{25}{121}}}=\frac{-3}{\sqrt{\frac{25}{121}}}=(-3):\sqrt{\frac{25}{121}}=(-3):\frac{\sqrt{25}}{\sqrt{121}}=(-3):\frac{5}{11}=(-3) \cdot \frac{11}{5}=\\ \\ = -\frac{33}{5}=-6\frac{3}{5}$

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Correct responses:

$Slope \ of \ \overline{IJ} = \underline{-\dfrac{7}{6} }$ , Length of $\overline{IJ}$ = $\underline{\sqrt{85} }$

$Slope \ of \ \overline{JK} = \underline {-\dfrac{2}{9}}$ , Length of $\overline{JK}$ = $\underline{\sqrt{85} }$

$Slope \ of \ \overline{KL} = \underline{-\dfrac{7}{6} }$ , Length of $\overline{KL}$ = $\underline{\sqrt{85}}$

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<h3>Methods used to find the slope and length of a line</h3>

The given coordinates of the vertices are;

K(-7, 0), J(2, -2), I(8, -9), L(-1, -7)

$Slope = \mathbf{\dfrac{y_2 - y_1}{x_2 - x_1}}$

$Length \ of \ line = \mathbf{ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}}$

Therefore;

$Slope \ of \ \mathbf{\overline{IJ}} = \dfrac{-2 - (-9)}{2 - 8} = \dfrac{7}{-6} = \underline{ -\dfrac{7}{6}}$

$Length \ of \ \mathbf{\overline{IJ}} = \sqrt{(2 - 8)^2 + (-2 - (-9))^2} = \underline{\sqrt{85}}$

$Slope \ of \ \mathbf{\overline{JK}} = \dfrac{0 - (-2)}{-7 - 2} = \dfrac{2}{-9} = \underline{-\dfrac{2}{9}}$

$Length \ of \ \mathbf{\overline{JK}} = \sqrt{(-7 - 2)^2 + (0 - (-2))^2} = \underline{ \sqrt{85}}$

$Slope \ of \ \mathbf{\overline{KL}} = \dfrac{0 - (-7)}{-7 - (-1)} = \dfrac{7}{-6} = \underline{ -\dfrac{7}{6}}$

$Length \ of \ \mathbf{\overline{KL} }= \sqrt{(-7 - (-1))^2 + (0 - (-7))^2} =\underline{ \sqrt{85}}$

$Slope \ of \ \mathbf{\overline{LI}} = \dfrac{(-9 - (-7))}{(8 - (-1))} = \dfrac{-2}{9} = \underline{-\dfrac{2}{9}}$

$Length \ of \ \mathbf{\overline{LI}} = \sqrt{(8 - (-1))^2 + (-9 - (-7))^2} = \underline{\sqrt{85}}$

Learn more about finding the slope and length of a line here:

brainly.com/question/11612395

brainly.com/question/14039630

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