1. 2.6
2. 0.9
3. 4.5
if you count the lines on the ruler, you can find the answer
She can break up the nine in to 3 three time and add it to 35
Answer:
C
Step-by-step explanation:
were is my brainliest, lol
Answer:
The property shown in matrix addition given is "Additive Inverse Property"
Step-by-step explanation:
First of all lets define what a matrix is.
A matrix is an array of rows and columns that consists of numbers. There are several types of matrices. The one in our question is a row matrix which consists of only one row.
There are several addition properties for matrices.
One of them is additive inverse property. The additive inverse of a matrix consists of the same elements but their signs are changed.
Additive inverse property states that the sum of a matrix and its additive inverse is a zero matrix.
![\left[\begin{array}{ccc}-6&15&-2\end{array}\right] + \left[\begin{array}{ccc}6&-15&2\end{array}\right] = \left[\begin{array}{ccc}0&0&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-6%2615%26-2%5Cend%7Barray%7D%5Cright%5D%20%2B%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D6%26-15%262%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%260%260%5Cend%7Barray%7D%5Cright%5D)
Hence,
The property shown in matrix addition given is "Additive Inverse Property"
So you can do this multiple ways, I'll do this the way that I think makes sense the l most easily.
Cos (0) = 1
Cos (pi/2)=0
Cos (pi) =-1
Cos (3pi/2)=0
Cos (2pi)=1
Now if you multiply the inside by 4, the graph oscillates more violently (goes up and down more in a shorter period).
But you can always reduce it.
Cos (0)= 1
Cos (4pi/2) = cos (2pi)=1
Cos (4pi) =Cos (2pi) =1 (Any multiple of 2pi ==1)
etc...
the pattern is that every half pi increase is now a full period as apposed to just a quarter of one. That's in theory.
Now that you know that, the identities of Cosine are another beast, but mathematically.
You have.
Cos (2×2t) = Cos^2 (2t)-Sin^2 (2t)
Sin^2 (t)=-Cos^2 (t)+1..... (all A^2+B^2=C^2)
Cos (2×2t) = Cos^2 (t)-(-Cos^2 (t)+1)
Cos (2×2t)= 2Cos^2 (2t) - 1
2Cos^2 (2t) -1= 2 (Cos^2(t)-Sin^2(t))^2 -1
(same thing as above but done twice because it's cos ^2 now)
convert sin^2
2Cos^2 (2t)-1 =2 (Cos^2 (t)+Cos^2 (t)-1)^2 -1
2 (2Cos^2(t)-1)^2 -1
2 (2Cos^2 (t)-1)(2Cos^2 (t)-1)-1
2 (4Cos^4 (t) - 2 (2Cos^2 (t))+1)-1
Distribute
8Cos^4 (t) -8Cos^2 (t) +1
Cos (4t) =8Cos^4-8Cos^2 (t)+-1
