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3 years ago

In order for an object to be a polygon it must have (choose the best answer) *

Mathematics

2 answers:

photoshop1234 [79]3 years ago

6 0

All sides must be a line segment

HACTEHA [7]3 years ago

3 0

Answer:

1. all side must intersect exactly two others sides but only at their endpoints

2. The sides must be non collinear and have a common endpoint

3. All sides must be a line segment

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Nilton acude a una institución privada que ofrece una tasa de interés del 0,5% mensual por el dinero que deposites, a un plazo f

jeka57 [31]

Answer:

hey, i might help in english?

Step-by-step explanation:

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3 years ago

Please solve this, will rate 5 stars and mark as STAR!

Nina [5.8K]

Answer:

$\boxed{5 \cdot \sqrt{2} \cdot \sqrt[6]{5} }$

Step-by-step explanation:

$\sqrt[3]{250} \cdot \sqrt{\sqrt[3]{10} }$

$\sqrt{\sqrt[3]{10} } \implies (10^\frac{1}{3} )^\frac{1}{2} =10^\frac{1}{6} =\sqrt[6]{10}$

$\therefore \sqrt{\sqrt[3]{10} }=\sqrt[6]{10}$

$\text{Solving }\sqrt[3]{250} \cdot \sqrt{\sqrt[3]{10} }$

$250=2 \cdot 5^3$

$\sqrt[3]{250}=\sqrt[3]{2\cdot 5^3}=5 \sqrt[3]{2}$

Once

$\sqrt[6]{2} \cdot \sqrt[6]{5} = \sqrt[6]{10}$

We have

$5 \sqrt[3]{2} \cdot \sqrt[6]{2} \cdot \sqrt[6]{5}$

We can proceed considering the common base of exponentials

$\sqrt[3]{2} \cdot \sqrt[6]{2} = 2^{\frac{1}{3}} \cdot 2^{\frac{1}{6} } = 2^{\frac{3}{6} } = 2^{\frac{1}{2} }=\sqrt{2}$

Therefore,

$5 \sqrt[3]{2} \cdot \sqrt[6]{2} \cdot \sqrt[6]{5} = 5 \cdot \sqrt{2} \cdot \sqrt[6]{5}$

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3 years ago

"I'm thinking of a fraction that is equivalent to 5/6

Liono4ka [1.6K]

Answer:

20/24

Step-by-step explanation:

let the variable be x and x-4

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3 years ago

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zloy xaker [14]

Answer:

Seven crore eighty thousands sixty-seven.

Step-by-step explanation:

70080067

Seven crore eighty thousands sixty-seven.

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3 years ago

Ollie borrowed $5000 for 3 years at a compound interest rate of 10%.

I am Lyosha [343]

I think Ollie payed $1,500 back.

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1 year ago

Read 2 more answers