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3 years ago

What is the sum of the measures of the exterior angles in a heptagon? Explain

Mathematics

2 answers:

sasho [114]3 years ago

7 0

The sum of exterior angles equals 360 degrees, which is the total degrees of a circle and all polygons

otez555 [7]3 years ago

6 0

Did you get the answer correct

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Help pls i really need help !!!

bixtya [17]

Answer:

that a tough one i'm guessing its longitude and latitude why not tell your teacher!

5 0

2 years ago

For the given triangles select the lengths of the sides.

aivan3 [116]

Answer:

v=3

u=6

x=2√2

y=2√2

Step-by-step explanation:

First triangle :

tanΦ=v/3√3

tan30=v/3√3

(√3)/3=v/(3√3)

Cross multiply

(√3)/3 x 3√3=v

(√3 x 3√3)/3=v

(3√9)/3=v

(3x3)/3=v

9/3=v

3=v

v=3

Sin30=v/u

0.5=3/u

Cross multiply

0.5xu=3

0.5u=3

Divide both sides by 0.5

0.5u/0.5=3/0.5

u=6

Second triangle :

sin45=x/4

Cross multiply

x=4 x sin45

x=4 x (√2)/2

x=2√2

Cos45=y/4

Cross multiply

4 x Cos45=y

4 x (√2)/2=y

(4√2)/2=y

2√2=y

y=2√2

5 0

2 years ago

Simplify

Diano4ka-milaya [45]

Answer:

$\huge\boxed{\sqrt[4]{16a^{-12}}=2a^{-3}=\dfrac{2}{a^3}}$

Step-by-step explanation:

$16=2^4\\\\a^{-12}=a^{(-3)(4)}=\left(a^{-3}\right)^4\qquad\text{used}\ (a^n)^m=a^{nm}\\\\\sqrt[4]{16a^{-12}}=\bigg(16a^{-12}\bigg)^\frac{1}{4}\qquad\text{used}\ a^\frac{1}{n}=\sqrt[n]{a}\\\\=\bigg(2^4(a^{-3})^4\bigg)^\frac{1}{4}\qquad\text{use}\ (ab)^n=a^nb^n\\\\=\bigg(2^4\bigg)^\frac{1}{4}\bigg[(a^{-3})^4\bigg]^\frac{1}{4}\qquad\text{use}\ (a^n)^m=a^{nm}\\\\=2^{(4)(\frac{1}{4})}(a^{-3})^{(4)(\frac{1}{4})}=2^1(a^{-3})^1=2a^{-3}\qquad\text{use}\ a^{-n}=\dfrac{1}{a^n}$