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lana [24]
2 years ago
12

I need help. I’ve been stuck

Mathematics
1 answer:
solniwko [45]2 years ago
3 0
Ok RYU? Isn’t it 110*?
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Complete the square to rewrite y=x^2-6x+14 in vertex form. Then state whether the vertex is a maximum or minimum and give is coo
Vladimir [108]

Answer:

3,5, Minimum

Step-by-step explanation:

y=x²-6x+14

y=(x²-2*3x +3²-3²)+14

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5 0
3 years ago
Lydia graphed ΔXYZ at the coordinates X (0, −4), Y (2, −3), and Z (2, −6). She thinks ΔXYZ is a right triangle. Is Lydia's asser
sammy [17]

Check the picture below.

so hmmm if Lydia is correct, then there's one angle in the triangle that is 90°, hmmm well, looking at the picture, we can pretty much forget about angle Z or Y, they're both acute, hmm how about angle X? is it 90°?

well, if angle X is indeed a right-angle, the lines XZ and XY are perpendicular, but are they?  if that's so then the slopes of XZ and XY are  negative reciprocal of each other, let's check

X(\stackrel{x_1}{0}~,~\stackrel{y_1}{-4})\qquad Z(\stackrel{x_2}{2}~,~\stackrel{y_2}{-6}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-6}-\stackrel{y1}{(-4)}}}{\underset{run} {\underset{x_2}{2}-\underset{x_1}{0}}} \implies \cfrac{-6 +4}{2 +0}\implies -1

now, the negative reciprocal of that will be

\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{-1\implies \cfrac{-1}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{-1}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{-1}\implies 1}}

well, let's see if XY has a slope is 1 then

X(\stackrel{x_1}{0}~,~\stackrel{y_1}{-4})\qquad Y(\stackrel{x_2}{2}~,~\stackrel{y_2}{-3}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-3}-\stackrel{y1}{(-4)}}}{\underset{run} {\underset{x_2}{2}-\underset{x_1}{0}}} \implies \cfrac{-3 +4}{2 +0}\implies \cfrac{1}{2}

OMG!!! Lydia needs to go get a nice Latte with cinnamon and to recheck her triangle.

4 0
2 years ago
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