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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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Step-by-step explanation:

16. Two parallel lines are cut by transversal. Angles with measures (6x+20)^{\circ} and (x+100)^{\circ} are alternate exterior angles. By alternate exterior angles, the measures of alternate exterior angles are the same:

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17. Two parallel lines are cut by transversal. Angles with measures (2x+12)^{\circ} and (3x-22)^{\circ} are alternate interior angles. By alternate interior angles, the measures of alternate interior angles are the same:

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18. Two parallel lines are cut by transversal. Angles with measures (6x-7)^{\circ} and (5x+10)^{\circ} are alternate exterior angles. By alternate interior angles, the measures of alternate exterior angles are the same:

6x-7=5x+10\\ \\6x-5x=10+7\\ \\x=17

Then

(6x-7)^{\circ}=(6\cdot 17-7)^{\circ}=95^{\circ}\\ \\(5x+10)^{\circ}=(5\cdot 17+10)^{\circ}=95^{\circ}

19. The diagram shows two complementary angles with measures 2x^{\circ} and 56^{\circ}. The measures of complementary angles add up to 90^{\circ}, then

2x+56=90\\ \\2x=90-56\\ \\2x=34\\ \\x=17

Hence,

2x^{\circ}=2\cdot 17^{\circ}=34^{\circ}

Check:

34^{\circ}+56^{\circ}=90^{\circ}

20. Angles \angle 1 and \angle 2 are vertical angles. By vertical angles theorem, vertical angles are congruent, so

m\angle 1=m\angle 2\\ \\5x+7=3x+15\\ \\5x-3x=15-7\\ \\2x=8\\ \\x=4

Hence,

m\angle 1=(5x+7)^{\circ}=(5\cdot 4+7)^{\circ}=27^{\circ}\\ \\m\angle 2=(3x+15)^{\circ}=(3\cdot 4+15)^{\circ}=27^{\circ}

21. \angle 5 and \angle 8 are supplementary. The measures of supplementary angles add up to 180^{\circ}, so

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Therefore,

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3 years ago
What is the equation of a circle with center (-4,7) radius 3?
miv72 [106K]

Answer:

a. (x + 4)² + (y – 7)² = 3²

Step-by-step explanation:

General equation for a circle is:

(x – h)² + (y – k)² = r²

h and k are the center (h, k), and r is the radius.

They want a center of (-4, 7) so h=-4 and k=7

They want a radius of 3 so r=3

plug it into the equation.

(x – h)² + (y – k)² = r²

(x – (-4))² + (y – (7))² = (3)²

(x + 4)² + (y – 7)² = 3²

7 0
4 years ago
Read 2 more answers
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