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3 years ago

Explain why a regular polygon cannot have an interior angle that is 40°.

Mathematics

1 answer:

Minchanka [31]3 years ago

8 0

Answer:

Explained below.

Step-by-step explanation:

The formula for the interior angle of any regular polygon is given as;

Interior Angle = 180(n - 2)/n

Where n is number of sides

We are told the interior angle is 40°

Thus;

180(n - 2)/n = 40

Cross multiply to get;

180n - 360 = 40n

180n - 40n = 360

140n = 360

n = 360/140

n = 2.57

Number of sides of a regular polygon cannot be in decimal nor can it have less than 3 sides.

Thus, a shape with interior angle of 40 cannot be a polygon.

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How many Solutions does this system have? (1 point)

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The given system of equation that is $2x+y=3$ and $6x=9-3y$ has infinite number of solutions.

Option -C.

Solution:

Need to determine number of solution given system of equation has.

$\begin{array}{l}{2 x+y=3} \\\\ {6 x=9-3 y}\end{array}$

Let us first bring the equation in standard form for comparison

$\begin{array}{l}{2 x+y-3=0} \\\\ {6 x+3 y-9=0}\end{array}$

$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

To check how many solutions are there for system of equations $a_{1} x+b_{1} y+c_{1}=0 \text{ and }a_{2} x+b_{2} y+c_{2}=0$ , we need to compare ratios of $\frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}} \text { and } \frac{c_{1}}{c_{2}}$

In our case,

$a_{1} = 2, b_{1}= 1\text{ and }c_{1}= -3$

$a_{2} = 6, b_{2} = 3,\text{ and }c_{2} = -9$

$\begin{array}{l}{\Rightarrow \frac{a_{1}}{a_{2}}=\frac{2}{6}=\frac{1}{3}} \\\\ {\Rightarrow \frac{b_{1}}{b_{2}}=\frac{1}{3}} \\\\ {\Rightarrow \frac{c_{1}}{c_{2}}=\frac{-3}{-9}=\frac{1}{3}} \\\\ {\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}=\frac{1}{3}}\end{array}$