Answer:
It should be A.
Step-by-step explanation:
When you're using exponents, anything that has an exponent of 0 will be 1.
You can use a calculator to confirm on your own.
Answer:
![\left[\begin{array}{cc}x&y\end{array}\right] * \left[\begin{array}{cc}3&1\\4&-2\end{array}\right] = \left[\begin{array}{cc}3x+4y&x-2y\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dx%26y%5Cend%7Barray%7D%5Cright%5D%20%2A%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%261%5C%5C4%26-2%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3x%2B4y%26x-2y%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
The general matrix representation for this transformation would be:
![\left[\begin{array}{cc}x&y\end{array}\right] * A = \left[\begin{array}{cc}3x+4y&x-2y\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dx%26y%5Cend%7Barray%7D%5Cright%5D%20%2A%20A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3x%2B4y%26x-2y%5Cend%7Barray%7D%5Cright%5D)
As the matrix A should have the same amount of rows as columns in the firs matrix and the same amount of columns as the result matrix it should be a 2x2 matrix.
![\left[\begin{array}{cc}x&y\end{array}\right] * \left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}3x+4y&x-2y\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dx%26y%5Cend%7Barray%7D%5Cright%5D%20%2A%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3x%2B4y%26x-2y%5Cend%7Barray%7D%5Cright%5D)
Solving the matrix product you have that the members of the result matrix are:
3x+4y = a*x + c*y
x - 2y = b*x + d*y
So the matrix A should be:
![\left[\begin{array}{cc}3&1\\4&-2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%261%5C%5C4%26-2%5Cend%7Barray%7D%5Cright%5D)
Answer:
x=nπ3, n∈I
Step-by-step explanation:
sin x + sin 5x = sin 2x + sin 4x
⇒⇒ 2 sin 3x cos 2x = 2 sin 3x cos x
⇒⇒ 2 sin 3x(cos 2x - cos x) = 0
⇒ sin 3x=0 ⇒ 3x=nπ ⇒ x=nπ3⇒ sin 3x=0 ⇒ 3x=nπ ⇒ x=nπ3 , n∈I, n∈I
or cos 2x−cos x=0 ⇒ cos 2x=cos xcos 2x-cos x=0 ⇒ cos 2x=cos x
⇒ 2x=2nπ±x ⇒ x=2nπ, 2nπ3⇒ 2x=2nπ±x ⇒ x=2nπ, 2nπ3 , n∈I, n∈I
But solutions obtained by x=2nπx=2nπ , n∈I, n∈I or x=2nπ3x=2nπ3 , n∈I, n∈I are all involved in x=nπ3x=nπ3 , n∈I
Answer:
a) 
b) 
c) 
Step-by-step explanation:
Given the curve
a) If the x-coordinate of P is 
, then the y-coordinate is 
so point P has coordinates 
If the x-coordinate of Q is 
, then the y-coordinate is 
so point Q has coordinates 
b) The gradient of the secant RQ is
c) If 
then the gradient 
Let's put it this way: If you plot a few non-x-intercept points and then draw a curvy line through them,you will not know if you got the x-intercepts even close to being correct. <span>The only way you can be sure of your x-intercepts is to set the quadratic equal to zero and solve. So its a matter of guessing from the pictures. Basicaly said, the calculator won't give you the exact result.</span>
