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mash [69]
3 years ago
9

When constructing an inscribed equilateral triangle, how many arcs will be drawn on the circle?

Mathematics
2 answers:
BlackZzzverrR [31]3 years ago
5 0

Answer:

it is c.5

Step-by-step explanation:

Lelechka [254]3 years ago
3 0

Answer:


Step-by-step explanation:

A. 3

You might be interested in
Consider the linear transformation T from V = P2 to W = P2 given by T(a0 + a1t + a2t2) = (2a0 + 3a1 + 3a2) + (6a0 + 4a1 + 4a2)t
Svet_ta [14]

Answer:

[T]EE=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]

Step-by-step explanation:

First we start by finding the dimension of the matrix [T]EE

The dimension is : Dim (W) x Dim (V) = 3 x 3

Because the dimension of P2 is the number of vectors in any basis of P2 and that number is 3

Then, we are looking for a 3 x 3 matrix.

To find [T]EE we must transform the vectors of the basis E and then that result express it in terms of basis E using coordinates and putting them into columns. The order in which we transform the vectors of basis E is very important.

The first vector of basis E is e1(t) = 1

We calculate T[e1(t)] = T(1)

In the equation : 1 = a0

T(1)=(2.1+3.0+3.0)+(6.1+4.0+4.0)t+(-2.1+3.0+4.0)t^{2}=2+6t-2t^{2}

[T(e1)]E=\left[\begin{array}{c}2&6&-2\\\end{array}\right]

And that is the first column of [T]EE

The second vector of basis E is e2(t) = t

We calculate T[e2(t)] = T(t)

in the equation : 1 = a1

T(t)=(2.0+3.1+3.0)+(6.0+4.1+4.0)t+(-2.0+3.1+4.0)t^{2}=3+4t+3t^{2}

[T(e2)]E=\left[\begin{array}{c}3&4&3\\\end{array}\right]

Finally, the third vector of basis E is e3(t)=t^{2}

T[e3(t)]=T(t^{2})

in the equation : a2 = 1

T(t^{2})=(2.0+3.0+3.1)+(6.0+4.0+4.1)t+(-2.0+3.0+4.1)t^{2}=3+4t+4t^{2}

Then

[T(t^{2})]E=\left[\begin{array}{c}3&4&4\\\end{array}\right]

And that is the third column of [T]EE

Let's write our matrix

[T]EE=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]

T(X) = AX

Where T(X) is to apply the transformation T to a vector of P2,A is the matrix [T]EE and X is the vector of coordinates in basis E of a vector from P2

For example, if X is the vector of coordinates from e1(t) = 1

X=\left[\begin{array}{c}1&0&0\\\end{array}\right]

AX=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]\left[\begin{array}{c}1&0&0\\\end{array}\right]=\left[\begin{array}{c}2&6&-2\\\end{array}\right]

Applying the coordinates 2,6 and -2 to the basis E we obtain

2+6t-2t^{2}

That was the original result of T[e1(t)]

8 0
3 years ago
What's the value of<br><img src="https://tex.z-dn.net/?f=2%20%5Ctimes%20750%20%20%2B%200%20%7B%7D%5E%7B7%7D%20" id="TexFormula1"
Elina [12.6K]

Step-by-step explanation:

2×750=1500 + 0^7. ( 0×0×0×0×0×0×0=0)

So,

1500 Answer

7 0
2 years ago
Read 2 more answers
Please answer i will give brainlest
Vitek1552 [10]

Answer:

here ya go

Step-by-step explanation:

1. 20

2.1793

3. 1943

im noy to sure on the bottom ones but use ctrl f in the notes you use for this or something and type in the "transacted" or something in the question and your answer should pop up

7 0
2 years ago
What is the domain of the function y=2 /X-6? EXCO o osx​
maks197457 [2]
I believe the answers for this is D
8 0
2 years ago
Read 2 more answers
What’s the correct answer for this?
zalisa [80]

Answer:

AP = 14

Step-by-step explanation:

According to secant-secant theorem

(CP)(PD)=(BP)(AP)

7×12=6×AP

AP = 84/6

AP = 14

3 0
3 years ago
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