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Marianna [84]
3 years ago
14

Next time, the rabbit from the previous problem start at 10 where on the number line can he be after the same three jumps?

Mathematics
1 answer:
Dimas [21]3 years ago
5 0

Answer:

3,5,7,9,11,13,15,17

Step-by-step explanation:

If rabbit starts at 0 the following are the possibilities

-4-2-1=-7

-4-2+1=-5

-4+2-1=-3

-4+2+1=-1

4-2-1=1

4-2+1=3

4+2-1=5

4+2+1=7

If the rabbit starts at 10 then

10-7=3

10-5=5

10-3=7

10-1=9

10+1=11

10+3=13

10+5=15

10+7=17

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Answer:

(x-1)!

Step-by-step explanation:

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3 years ago
Indicate the equation of the given line in standard form. The line that is the perpendicular bisector of the segment whose endpo
noname [10]
Well the line that bisects RS, will cut RS in two equal halves, therefore, that line will cut RS perpendicularly at the midpoint of RS.

now, what the dickens is the midpoint of RS anyway?

\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
R&({{ -1}}\quad ,&{{ 6}})\quad 
%  (c,d)
S&({{ 5}}\quad ,&{{ 5}})
\end{array}\qquad
%   coordinates of midpoint 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)
\\\\\\
\left( \cfrac{5-1}{2}~~,~~\cfrac{5+6}{2} \right)\implies \stackrel{midpoint}{\left(2~~,~~\frac{11}{2}  \right)}

so, we know that perpendicular line, will have to go through (2, 11/2)

now, a perpendicular line to RS, will have a negative reciprocal slope to it.  Well, what is the slope of RS anyway?

\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%   (a,b)
&({{ -1}}\quad ,&{{ 6}})\quad 
%   (c,d)
&({{ 5}}\quad ,&{{ 5}})
\end{array}
\\\\\\
% slope  = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{5-6}{5-(-1)}\implies \cfrac{5-6}{5+1}\implies -\cfrac{1}{6}

and let's check the reciprocal negative of that,

\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad -\cfrac{1}{6}\\\\
slope=-\cfrac{1}{{{ 6}}}\qquad negative\implies  +\cfrac{1}{{{ 6}}}\qquad reciprocal\implies + \cfrac{{{ 6}}}{1}\implies 6

so, then, what's is the equation of a line whose slope is 6, and goes through 2, 11/2?

\bf \begin{array}{lllll}
&x_1&y_1\\
%   (a,b)
&({{ 2}}\quad ,&{{ \frac{11}{2}}})
\end{array}
\\\\\\
% slope  = m
slope = {{ m}}= \cfrac{rise}{run} \implies 6
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-\cfrac{11}{2}=6(x-2)
\\\\\\
y-\cfrac{11}{2}=6x-12\implies -6x+y=-12+\cfrac{11}{2}\implies \stackrel{\textit{standard form}}{-6x+y=-\cfrac{13}{2}}
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3 years ago
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ehidna [41]
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Answer:

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prohojiy [21]

Answer:

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