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3 years ago

Is the centroid of a triangle equidistant from the three vertices?

Mathematics

1 answer:

Xelga [282]3 years ago

4 0

The point that is equidistant to all sides of a triangle is called the incenter.

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9 (1/3)^n-1 what is the tenth term

NARA [144]

Answer:

$a_{10}=\frac{1}{2187}$

Step-by-step explanation:

Simply plug in 10 for n

9(1/3)¹⁰⁻¹

9(1/3)⁹

9(1/19683)

9/19683

$a_{10}=\frac{1}{2187}$

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3 years ago

What are the multiples of 4

N76 [4]

1 and 2 are the factors its a compatable number

8 0

3 years ago

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Solve sin 0 + 1 = cos20 on the interval 0 ≤ 0 < 2pi. Show work please!

yan [13]

Answer:

$\theta=\frac{\pi}{2},\frac{3\pi}{2}\frac{2\pi}{3}\frac{4\pi}{3}$

Step-by-step explanation:

You need 2 things in order to solve this equation: a trig identity sheet and a unit circle.

You will find when you look on your trig identity sheet that

$cos(2\theta)=1-2sin^2(\theta)$

so we will make that replacement, getting everything in terms of sin:

$sin(\theta)+1=1-2sin^2(\theta)$

Now we will get everything on one side of the equals sign, set it equal to 0, and solve it:

$2sin^2(\theta)+sin(\theta)=0$

We can factor out the sin(theta), since it's common in both terms:

$sin(\theta)(2sin(\theta)+1)=0$

Because of the Zero Product Property, either

$sin(\theta)=0$ or

$2sin(\theta)+1=0$

Look at the unit circle and find which values of theta have a sin ratio of 0 in the interval from 0 to 2pi. They are:

$\theta=\frac{\pi}{2},\frac{3\pi}{2}$

The next equation needs to first be solved for sin(theta):

$2sin(\theta)+1=0$ so

$2sin(\theta)=-1$ and

$sin(\theta)=-\frac{1}{2}$

Go back to your unit circle and find the values of theta where the sin is -1/2 in the interval. They are:

$\theta=\frac{2\pi}{3},\frac{4\pi}{3}$

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2 years ago

Can someone help me with this?

Lerok [7]

Answer:

1 is C and 2 is B

Step-by-step explanation:

4 0

2 years ago

Estimate how many 12 inch rulers will be about the same length as a window

Shkiper50 [21]

2-3 12 INCH RULERS
sorry 4 caps

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3 years ago

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