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ratelena [41]
2 years ago
14

What are the coordinates of the terminal point for 0 = 4pi/3

Mathematics
1 answer:
nasty-shy [4]2 years ago
6 0

Answer:

I hope this helps. There is a link

b

So what is that distance well we know that to get to negative 4 pi over 3 we took negative pi and travelled pi over 3.. So this is just pi over 3.. So our reference angle would be pi over 3.

Step-by-step explanation:

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Please help! Will give Brainliest!
kkurt [141]

Answer:

  1. g(x) = 2|x|
  2. g(x) = -2|x|
  3. g(x) = -2|x| -3
  4. g(x) = -2|x-1| -3

Step-by-step explanation:

1) Vertical stretch is accomplished by multiplying the function value by the stretch factor. When |x| is stretched by a factor of 2, the stretched function is ...

  g(x) = 2|x|

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2) Reflection over the x-axis means each y-value is replaced by its opposite. This is accomplished by multiplying the function value by -1.

  g(x) = -2|x|

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3) As you know from when you plot a point on a graph, shifting it down 3 units subtracts 3 from the y-value.

  g(x) = -2|x| -3

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4) A right-shift by k units means the argument of the function is replaced by x-k. We want a right shift of 1 unit, so ...

  g(x) = -2|x -1| -3

7 0
3 years ago
Solve the system by using a matrix equation.<br> --4x - 5y = -5<br> -6x - 8y = -2
evablogger [386]

Answer:

Solution : (15, - 11)

Step-by-step explanation:

We want to solve this problem using a matrix, so it would be wise to apply Gaussian elimination. Doing so we can start by writing out the matrix of the coefficients, and the solutions ( - 5 and - 2 ) --- ( 1 )

\begin{bmatrix}-4&-5&|&-5\\ -6&-8&|&-2\end{bmatrix}

Now let's begin by canceling the leading coefficient in each row, reaching row echelon form, as we desire --- ( 2 )

Row Echelon Form :

\begin{pmatrix}1\:&\:\cdots \:&\:b\:\\ 0\:&\ddots \:&\:\vdots \\ 0\:&\:0\:&\:1\end{pmatrix}

Step # 1 : Swap the first and second matrix rows,

\begin{pmatrix}-6&-8&-2\\ -4&-5&-5\end{pmatrix}

Step # 2 : Cancel leading coefficient in row 2 through R_2\:\leftarrow \:R_2-\frac{2}{3}\cdot \:R_1,

\begin{pmatrix}-6&-8&-2\\ 0&\frac{1}{3}&-\frac{11}{3}\end{pmatrix}

Now we can continue canceling the leading coefficient in each row, and finally reach the following matrix.

\begin{bmatrix}1&0&|&15\\ 0&1&|&-11\end{bmatrix}

As you can see our solution is x = 15, y = - 11 or (15, - 11).

4 0
3 years ago
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