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Vsevolod [243]
3 years ago
15

Help me plss need help with theses three

Mathematics
1 answer:
fiasKO [112]3 years ago
4 0
interior angle of a regular 18-gon.
It is easier to calculate the exterior angle of a regular polygon of n-sides (n-gon) by the relation
exterior angle = 360/n
For a 18-gon, n=18, so exterior angle = 360/18=20 °
The value of each interior angle is therefore the supplement, or
Interior angle = 180-20=160 degrees.

Naming of a 9-gon
A polygon with 9 vertices is called a nonagon (in English) or enneagon (French ennéagone, but the English version is sometimes used)
You had a good start with the correct answer.

Exterior angle of a 15-gon
The exterior angle of a 15-gon can be calculated using the relation given in the first paragraph, namely
Exterior angle = 360/15=24 degrees
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Josslyn has nickels and dimes in her pocket. The number of nickels is three more than seven times the number of dimes let d repr
vfiekz [6]
7d+3Explanation

to solve this we need to translate into math terms, so

Step 1

a) let d represents the number of dimes

let n represents the number of nickles

so

re write the expressions

\begin{gathered} number\text{ of dimes=d} \\ seven\text{ times the number of dimes = 7d} \\  \end{gathered}

The number of nickels is three more than seven times the number of dimes in other words you have to add 7 to seven times the number of dimes to obtain the number of nickles

hence

n=7d+3

therefore , the expression for the number of nickles is

7d+3

I hope this helps you

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1 year ago
ajwad went apple picking.He picked a total of 84 apples and brought them all to school.If there are 21 students,how many apples
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I'm pretty sure the answer is 4 because, 4*21= 84.
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Step-by-step explanation:

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(1+cos2x)/(1-cos2x) = cot^2x
sesenic [268]

We will turn the left side into the right side.

\dfrac{1 + \cos 2x}{1 - \cos2x} = \cot^2 x

Use the identity:

\cos 2x = \cos^2 x - \sin^2 x

\dfrac{1 + \cos^2 x - \sin^2 x}{1 - ( \cos^2 x - \sin^2 x)} = \cot^2 x

\dfrac{1 - \sin^2 x + \cos^2 x }{1 - \cos^2 x + \sin^2 x} = \cot^2 x

Now use the identity

\sin^2 x + \cos^2 x = 1 solved for sin^2 x and for cos^2 x.

\dfrac{\cos^2 x + \cos^2 x }{\sin^2 x + \sin^2 x} = \cot^2 x

\dfrac{2\cos^2 x}{2\sin^2 x} = \cot^2 x

\dfrac{\cos^2 x}{\sin^2 x} = \cot^2 x

\cot^2 x = \cot^2 x


8 0
3 years ago
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