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Answers

Answer 1

Answer:

Answer:

Both Jules' and Lauren's equations are correct because they have slopes that are the negative reciprocal of the slope of the given line, making them perpendicular to the given line.

Step-by-step explanation:

Let's reevaluate the equations based on the corrected given line equation:

$\sf y - 2 = (1)/(5)(x - 3)$

The given line equation is in point-slope form: $\sf \boxed{\sf y - y_1 = m(x - x_1)}$ , where m is the slope.

Given line equation: $\sf y - 2 = (1)/(5)(x - 3)$

While comparing, we get

$\textsf{The\underline{ slope (m) }of the given line is }(1)/(5)$

For a line to be perpendicular to the given line, its slope must be the negative reciprocal of the slope of the given line.

The negative reciprocal of $\sf (1)/(5)$ is $\sf -5.$

Now let's check the slopes of the equations provided by Jules and Lauren:

1. Jules' equation: $\sf y = -5x + 1$

The slope of Jules' equation is -5, which matches the negative reciprocal of the slope of the given line.

2. Lauren's equation: $\sf y = -5x + 7$

The slope of Lauren's equation is also -5, which again matches the negative reciprocal of the slope of the given line.

Both Jules' and Lauren's equations have a slope of -5, which is the negative reciprocal of the slope of the given line $(1)/(5)$ .

Therefore, both equations are correct and satisfy the condition of being perpendicular to the given line $\sf y - 2 = (1)/(5)(x - 3)$

Answer 2

Answer:

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above

$y-2=\stackrel{\stackrel{m}{\downarrow }}{\cfrac{1}{5}}(x-3)\qquad \impliedby \qquad \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\n \cline{1-1} \n y-y_1=m(x-x_1) \n\n \cline{1-1} \end{array} \n\n[-0.35em] ~\dotfill$

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

so ANY line that is perpendicular to that equation above, will have a slope of -5, so any of these are all perpendicular to it

$\begin{array}{llll} \stackrel{ Jules }{y=-5x+1} \n\n\n \stackrel{ Lauren }{y=-5x+7} \n\n\n y=-5x+999999999 \n\n\n y=-5x-93789 \end{array}\hspace{5em} \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\n \cline{1-1} \n y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \n\n \cline{1-1} \end{array}$

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Answers

4 dollars.

Remember

1-40 is is rounded back to one
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$4.00 would be the answer

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what does z equal

Answers

add 2.2 to each side so 0.3z = 5.2 - 0.5z -0.8z
add 0.5z to each side so 0.8z = 5.2 - 0.8z
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Find the value of x for which abcd must be a parallelogram

Answers

Answer:

x = 7

Step-by-step explanation:

3x - 17 = 25 - 3x

Get the x's on one side and the integers on the other

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F n and m are integers and n2 m2 is even, which of the following is impossible?

Answers

So, I think that you mean that

n²+m² is even? (the "plus" sign is missing, but I think this is correct).

this can only happen if either:

both n and m are even
or
both n and m are odd

That's because a sum of two expressions is even if either both or none of the expressions summed is even, and a power of a number is even only if the number is even too.

are those some of the options?

yes!
n and m are even,- possible!
n and m are odd, -possible!
n+m is even, - possible; it would be the case if either both n and m are even or both n and m are odd n+m is odd,

so the option we'll choose is:

none of these are impossible

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Answers

First set the fractions into improper fractions...
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