<em>The distance the length of a segment with endpoints Y(2, 8) and Z(-2, 5) is 5 units</em>
<h2>Explanation:</h2>
Endpoints of a Line segments are places where they end or stop. Line segments are named after their endpoints. In this case, those endpoints are Y and Z, so the line segment would be:
![\overline{YZ} \ or \ \overline{ZY}](https://tex.z-dn.net/?f=%5Coverline%7BYZ%7D%20%5C%20or%20%5C%20%5Coverline%7BZY%7D)
To find the length of this segment with endpoints Y(2, 8) and Z(-2, 5), let's use the Distance Formula:
![d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%28x_%7B2%7D-x_%7B1%7D%29%5E2%2B%28y_%7B2%7D-y_%7B1%7D%29%5E2%7D)
![Let's \ write: \\ \\ Y(x_{1},y_{1}) \rightarrow Y(2, 8) \\ \\ Z(x_{2},y_{2}) \rightarrow (-2, 5) \\ \\ \\ Substituting: \\ \\ d=\sqrt{(-2-2))^2+(5-8)}^2 \\ \\ d=\sqrt{(-4)^2+(-3)^2} \\ \\ d=\sqrt{16+9} \\ \\ d=\sqrt{25} \\ \\ \boxed{d=5}](https://tex.z-dn.net/?f=Let%27s%20%5C%20write%3A%20%5C%5C%20%5C%5C%20Y%28x_%7B1%7D%2Cy_%7B1%7D%29%20%5Crightarrow%20Y%282%2C%208%29%20%5C%5C%20%5C%5C%20Z%28x_%7B2%7D%2Cy_%7B2%7D%29%20%5Crightarrow%20%28-2%2C%205%29%20%5C%5C%20%5C%5C%20%5C%5C%20Substituting%3A%20%5C%5C%20%5C%5C%20d%3D%5Csqrt%7B%28-2-2%29%29%5E2%2B%285-8%29%7D%5E2%20%5C%5C%20%5C%5C%20d%3D%5Csqrt%7B%28-4%29%5E2%2B%28-3%29%5E2%7D%20%5C%5C%20%5C%5C%20d%3D%5Csqrt%7B16%2B9%7D%20%20%5C%5C%20%5C%5C%20d%3D%5Csqrt%7B25%7D%20%5C%5C%20%5C%5C%20%5Cboxed%7Bd%3D5%7D)
Finally, <em>the distance the length of a segment with endpoints Y(2, 8) and Z(-2, 5) is 5 units</em>
<h2>Learn more:</h2>
Distance Formula: brainly.com/question/10134840
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4 has the most. Sorry if it’s wrong, good day to you
C) !!!
In ratios, 'things go with the same things', so recliners have to be both in the numerator or denominator. The 8 is misleading. That's the total, and the problem is all about recliners and couches only
Answer:
![V =\frac{E}{IT}](https://tex.z-dn.net/?f=V%20%3D%5Cfrac%7BE%7D%7BIT%7D)
Step-by-step explanation:
You can isolate the "V" variable by dividing by IT on both sides:
![\frac{E}{IT} =\frac{VIT}{IT}](https://tex.z-dn.net/?f=%5Cfrac%7BE%7D%7BIT%7D%20%3D%5Cfrac%7BVIT%7D%7BIT%7D)
On the left, the IT from the top and bottom cancel, leaving you with just V:
![V =\frac{E}{IT}](https://tex.z-dn.net/?f=V%20%3D%5Cfrac%7BE%7D%7BIT%7D)