Y = (x+3)^2 - 5 here's the answer hope it helps
Answer:
A_16x20=320
Step-by-step explanation:
Esto se puede hacer fácilmente de 2 formas.
1) 20 se puede dividir en un múltiplo de 10 y luego multiplicar
16 x 20 =16 x 10 x 2 = 160 x 2 = 320
2) Uso de cero.
Cualquier cosa multiplicada por cero da un cero y luego multiplicar el otro número da el resultado.
16 x 20 = (16 x 2) * 10 = (32) 10 = 320
Así que separamos el cero añadiéndole un 1, lo que lo convierte en 10 y obtenemos el mismo resultado.
This can be easily done in 2 ways .
1) 20 can be broken into a multiple of 10 and then multiplied
16 x 20 =16 x10 x 2=160x 2= 320
2) Use of zero.
Anything multiplied with zero gives a zero and then multiplying the other number gives the result.
16 x 20 = (16x 2) *10= (32)10= 320
So we separated the zero by adding a 1 to it thus making it 10 and we get the same result.
Answer:
s a vertex is a corner so s
First, we have to make sure that the number of columns in the first matrix is equal to the number of rows in the second matrix.
![\left[\begin{array}{cc}1&-3&2&0\\\end{array}\right] * \left[\begin{array}{ccc}2&3&4\\1&2&3\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26-3%262%260%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%2A%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%263%264%5C%5C1%262%263%5Cend%7Barray%7D%5Cright%5D%20)
Since this is true, we can continue to solve the problem.
To multiply two matrices, multiply each row element in the first matrix by each column element in the second matrix. For example:
1*2 = 2
-3*1=-3
Then we add them to get our new matrix element.
-3+2=
-1Then we move to the next column of the second matrix.
1*3=3
-3*2=-6
-6+3=
-3Then the final column of the second matrix.
1*4=4
-3*3=-9
-9+4=-5
Our matrix so far:
![\left[\begin{array}{ccc}-1&-3&-5\\x&x&x\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%26-3%26-5%5C%5Cx%26x%26x%5Cend%7Barray%7D%5Cright%5D%20)
We do the same for the bottom row of the first matrix.
<em>First Column</em>
2*2=4
0*1=0
4+0=
4<em>Second Column
</em>2*3=6
0*2=0
6+0=
6
<em>Third Column</em>
2*4=8
0*3=0
8+0=
8Our final matrix is:
![\left[\begin{array}{ccc}-1&-3&-5\\4&6&8\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%26-3%26-5%5C%5C4%266%268%5Cend%7Barray%7D%5Cright%5D)
:)
Answer:
Is it 12?Yes,I think it's 12.