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cupoosta [38]
3 years ago
5

What is the opisite of 47 helppp!!!!

Mathematics
2 answers:
tensa zangetsu [6.8K]3 years ago
8 0

Answer:

-47

Step-by-step explanation:

REY [17]3 years ago
3 0
The answer would be -47
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What is the area, in square centimeters, of the shape below?
Karolina [17]

Answer:

21.46 cm squared

Step-by-step explanation:

3.8x5.8=21.46

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*EXTRA PTS* pls answer all 3 :)
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Answer:

5, x = 0,  y = -2

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7,  X = -2, Y = -2

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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sashaice [31]

Answer:

you need to show the shape or somet or else I can't rlly help

7 0
3 years ago
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Find two integers whose sum is 5 and product is 4
gregori [183]

Answer:

1 and 4

Step-by-step explanation:

1+4=5

1 x 4=4

6 0
4 years ago
Read 2 more answers
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