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3 years ago

The interior angles of a hexagon are (2x+17)0, (3x - 25)0, (2x+49)0,

Mathematics

1 answer:

EleoNora [17]3 years ago

7 0

Answer: a) x= 68, b) 108°, c) 72°, d) 255°.

Step-by-step explanation:

Since we have given that

The interior angles of hexagon:

(2x+17), (3x - 25), (2x+49), (x+40), (4x-17) and (3x - 4).

So, it becomes :

$2x+17+3x-25+2x+49+x+40+4x-17+3x-4=1080\\\\15x+60=1080\\\\15x=1080-60\\\\15x=1020\\\\x=68$

ii. Find the smallest interior angle of the quadrilateral.

$x+40=68+40=108^\circ$

iii. Find the largest exterior angle of the quadrilateral.

$180-108=72^\circ$

iv. Find the largest interior angle of the quadrilateral.

$4x-17=4\times 68-17=255^\circ$

Hence, a) x= 68, b) 108°, c) 72°, d) 255°.

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3 years ago

Given the circle with the equation (x + 1)2 + y2 = 36, determine the location of each point with respect to the graph of the cir

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$\bf \textit{equation of a circle}\\\\ 
(x- h)^2+(y- k)^2= r^2
\qquad 
center~~(\stackrel{}{ h},\stackrel{}{ k})\qquad \qquad 
radius=\stackrel{}{ r}\\\\
-------------------------------\\\\
(x+1)^2+y^2=36\implies [x-(\stackrel{h}{-1})]^2+[y-\stackrel{k}{0}]^2=\stackrel{r}{6^2}~~~~
\begin{cases}
\stackrel{center}{(-1,0)}\\
\stackrel{radius}{6}
\end{cases}$

so, that's the equation of the circle, and that's its center, any point "ON" the circle, namely on its circumference, will have a distance to the center of 6 units, since that's the radius.

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
(\stackrel{x_1}{-1}~,~\stackrel{y_1}{0})\qquad 
A(\stackrel{x_2}{-1}~,~\stackrel{y_2}{1})\qquad \qquad 
% distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
\stackrel{distance}{d}=\sqrt{[-1-(-1)]^2+(1-0)^2}\implies d=\sqrt{(-1+1)^2+1^2}
\\\\\\
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well, the distance from the center to A is 1, namely is "inside the circle".

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}
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(\stackrel{x_1}{-1}~,~\stackrel{y_1}{0})\qquad 
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\\\\\\
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notice, the distance to B is exactly 6, and you know what that means.

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
(\stackrel{x_1}{-1}~,~\stackrel{y_1}{0})\qquad 
C(\stackrel{x_2}{4}~,~\stackrel{y_2}{-8})
\\\\\\
\stackrel{distance}{d}=\sqrt{[4-(-1)]^2+[-8-0]^2}\implies d=\sqrt{(4+1)^2+(-8)^2}
\\\\\\
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notice, C is farther than the radius 6, meaning is outside the circle, hiking about on the plane.

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3 years ago

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