All categories

3 years ago

The sum of the measure of the interior angles of a polygon is 1980°. What polygon is it? 10-gon 11-gon 12-gon 13-gon

Mathematics

1 answer:

tiny-mole [99]3 years ago

3 0

Answer:

13-gon

Step-by-step explanation:

The sum of the interior angles of a polygon with n sides is:

S = 180° (n − 2)

If S = 1980°, then:

1980° = 180° (n − 2)

11 = n − 2

n = 13

You might be interested in

Could someone help me, please?

k0ka [10]

We can rewrite the expression under the radical as

$\dfrac{81}{16}a^8b^{12}c^{16}=\left(\dfrac32a^2b^3c^4\right)^4$

then taking the fourth root, we get

$\sqrt[4]{\left(\dfrac32a^2b^3c^4\right)^4}=\left|\dfrac32a^2b^3c^4\right|$

Why the absolute value? It's for the same reason that

$\sqrt{x^2}=|x|$

since both $(-x)^2$ and $x^2$ return the same number $x^2$ , and $|x|$ captures both possibilities. From here, we have

The absolute values disappear on all but the $b$ term because all of $\dfrac32$ , $a^2$ and $c^4$ are positive, while $b^3$ could potentially be negative. So we end up with

$\dfrac32a^2\left|b^3\right|c^4=\dfrac32a^2|b|^3c^4$

3 0

3 years ago

Older headstones are often difficult to read due to weathering that has worn away their letters. This is an example of which of

zavuch27 [327]

Answer:

Abrasion

Step-by-step explanation:

Abrasion is the process of scraping or wearing away something. If weathering has worn away the letters, it has been abrasion weathering.

3 0

3 years ago

Let <img src="https://tex.z-dn.net/?f=i" id="TexFormula1" title="i" alt="i" align="absmiddle" class="latex-formula"> be the imag

VLD [36.1K]

$Hey~freckledspots!\\----------------------$

$We~will~solve~for~i^{425}!$

$Rule~of~exponent: a^{b + c} = a^ba^c\\Apply:~i^{425}~=~i^{424}i\\ \\Rule~of~exponent: a^{bc} = (a^{b})^c\\Apply: i^{424} = i(i^2)^{212} \\\\Rule~of~imaginary~number: i^2 = -1\\Apply: i(i^2)^{212} = -1^{212}i\\\\Rule~of~exponent~if~n~is~even: -a^n = a^n\\Apply: -1^{212}i = 1^{212}i\\\\Simplify: 1^{212}i = 1i\\Multiply: 1i * 1 = i\\----------------------\\$

$Now~let's~solve~1^{14}!\\\\Rule~of~exponent: a^{b + c} = a^ba^c\\Apply: i^{14} = (i^2)^7\\\\Rule~of~imaginary~number: i^2 = -1\\Apply: (i^2)^7 = -1^7\\\\Rule~of~exponent~if~n~is~odd: (-a)^n = -a^n\\Apply: -1^7 = -1^7\\\\Simplify: -1^7 = -1\\----------------------\\Now,~we~have: i-1+i^{-14}+i^{44}\\----------------------$

$Now~lets~solve~i^{-14}\\\\Rule~of~exponent: a^{-b} = \frac{1}{a^b} \\Apply: i^{-14} = \frac{1}{i^{14}} \\\\Rule~of~exponent: a^{bc} = (a^b)^c\\Apply: \frac{1}{i^{14}} = \frac{1}{(i^2)^7}\\ \\Rule~of~imagianry~number: i^2 = -1\\Apply: \frac{1}{(i^2)^7} = \frac{1}{-1^7} \\\\Simplify: \frac{1}{-1^7} = \frac{1}{-1} \\\\Rule~of~fractions: \frac{a}{-b} = -\frac{a}{b} \\Apply: \frac{1}{-1} = -\frac{1}{1} = -1\\----------------------\\Now,~we~have: i-1-1+i^44\\----------------------$

$Now~let's~solve~i^{44}!\\\\Rule~of~exponent: a^{bc} = (a^b)^c\\Apply: i^{44} = (i^2)^{22}\\\\Rule~of~imaginary~numbers: i^2 = -1\\Apply: (i^2)^{22} = -1^{22}\\\\Rule~of~exponent~if~n~is~even: (-a)^n = a^n\\Apply: -1^{22} = 1^{22}\\\\Simplify: 1^{22} = 1\\----------------------\\Now,~we~have~i-1-1+1\\----------------------$