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damaskus [11]
3 years ago
11

How do you solve all of these ?

Mathematics
1 answer:
Flura [38]3 years ago
4 0
5)

\bf \begin{array}{lllll}
&x_1&y_1\\
%   (a,b)
&({{ 5}}\quad ,&{{ -3}})\quad 
%   (c,d)
\end{array}
\\\\\\
% slope  = m
slope = {{ m}}= \cfrac{rise}{run} \implies -1
\\\\\\
% point-slope intercept
y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-(-3)=-1(x-5)
\\
\left. \qquad   \right. \uparrow\\
\textit{point-slope form}
\\\\\\
y+3=-x+5\implies y=-x+2

6)

\bf \begin{array}{lllll}
&x_1&y_1\\
%   (a,b)
&({{ 2}}\quad ,&{{ 1}})\quad 
%   (c,d)
\end{array}
\\\\\\
% slope  = m
slope = {{ m}}= \cfrac{rise}{run} \implies 3
\\\\\\
% point-slope intercept
y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-1=3(x-2)
\\
\left. \qquad   \right. \uparrow\\
\textit{point-slope form}
\\\\\\
y-1=3x-6\implies y=3x-5

7)

\bf \begin{array}{lllll}
&x_1&y_1\\
%   (a,b)
&({{ 5}}\quad ,&{{ 2}})\quad 
%   (c,d)
\end{array}
\\\\\\
% slope  = m
slope = {{ m}}= \cfrac{rise}{run} \implies 0
\\\\\\
% point-slope intercept
y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-2=0(x-5)
\\
\left. \qquad   \right. \uparrow\\
\textit{point-slope form}
\\\\\\
y-2=0\implies y=2

8)

\bf \begin{array}{lllll}
&x_1&y_1\\
%   (a,b)
&({{ -2}}\quad ,&{{ 0}})\quad 
%   (c,d)
\end{array}
\\\\\\
% slope  = m
slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{5}{2}
\\\\\\
% point-slope intercept
y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-0=\cfrac{5}{2}(x-(-2))
\\
\left. \qquad   \right. \uparrow\\
\textit{point-slope form}
\\\\\\
y=\cfrac{5}{2}(x+2)\implies y=\cfrac{5}{2}x+5

9)

\bf \begin{array}{lllll}
&x_1&y_1\\
%   (a,b)
&({{ -2}}\quad ,&{{ 0}})\quad 
%   (c,d)
\end{array}
\\\\\\
% slope  = m
slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{5}{2}
\\\\\\
% point-slope intercept
y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-0=\cfrac{5}{2}(x-(-2))
\\
\left. \qquad   \right. \uparrow\\
\textit{point-slope form}
\\\\\\
y=\cfrac{5}{2}(x+2)\implies y=\cfrac{5}{2}x+5

10)

\bf \begin{array}{lllll}
&x_1&y_1\\
%   (a,b)
&({{-2}}\quad ,&{{ -2}})\quad 
%   (c,d)
\end{array}
\\\\\\
% slope  = m
slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{3}{2}
\\\\\\
% point-slope intercept
y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-(-2)=\cfrac{3}{2}(x-(-2))
\\
\left. \qquad   \right. \uparrow\\
\textit{point-slope form}
\\\\\\
y+2=\cfrac{3}{2}(x+2)\implies y+2=\cfrac{3}{2}x+3\implies y=\cfrac{3}{2}x+1
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Answer:

46degrees

Step-by-step explanation:

From the question, we are told that both triangles are similar, hence;

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Since <D = <F = 67

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<E = 180 - 134

<E = 46degrees

Since <E = <S, hence the measure of <S  is 46degrees

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