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2 years ago

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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3 years ago

The ratio of the sides of two squares is 3:1. what is the ratio of their perimeters?

inessss [21]

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hoped this helped

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3 years ago

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2 years ago

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

y=(x²-2*3x+3²)+14-9

y=(x-3)²+5

a=1

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q=5

a>0 ⇒ vertex is minimum

V(p,q) ⇒ V(3;5)

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