The product of A and B matrix will be option D; BA = ![\left[\begin{array}{ccc}16&10&39\\-32&-30&-18\\-6&-6&-19\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D16%2610%2639%5C%5C-32%26-30%26-18%5C%5C-6%26-6%26-19%5Cend%7Barray%7D%5Cright%5D)
.
<h3>What is the multiplication of matrix?</h3>
If ![A = \left[\begin{array}{ccc}-2&-4&2\\5&1&5\\-1&-3&-4\end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%26-4%262%5C%5C5%261%265%5C%5C-1%26-3%26-4%5Cend%7Barray%7D%5Cright%5D)
and ![B = \left[\begin{array}{ccc}2&3&-5\\5&-4&2\\-1&-1&3\end{array}\right]](https://tex.z-dn.net/?f=B%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%263%26-5%5C%5C5%26-4%262%5C%5C-1%26-1%263%5Cend%7Barray%7D%5Cright%5D)
then,
BA = ![\left[\begin{array}{ccc}2&3&-5\\5&-4&2\\-1&-1&3\end{array}\right] \left[\begin{array}{ccc}-2&-4&2\\5&1&5\\-1&-3&-4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%263%26-5%5C%5C5%26-4%262%5C%5C-1%26-1%263%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%26-4%262%5C%5C5%261%265%5C%5C-1%26-3%26-4%5Cend%7Barray%7D%5Cright%5D)
BA =
Hence, the product of A and B matrix will be option D; BA = .
Learn more about matrix here;
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Answer:
4
Step-by-step explanation:
Answer:
a
Step-by-step explanation:
Answer:
25
Explanation:
By Using Pythagoras Theorem,
h^2=p^2+b^2
or, x^2=(24)^2+7^2
or, x^2=576+49
or, x^2=625
or, x=√625
•°• x=25
Answer:
Vertical A @ x=3 and x=1
Horizontal A nowhere since degree on top is higher than degree on bottom
Slant A @ y=x-1
Step-by-step explanation:
I'm going to look for vertical first:
I'm going to factor the bottom first: (x-3)(x-1)
So we have possible vertical asymptotes at x=3 and at x=1
To check I'm going to see if (x-3) is a factor of the top by plugging in 3 and seeing if I receive 0 (If I receive 0 then x=3 gives me a hole)
3^3-5(3)^2+4(3)-25=-31 so it isn't a factor of the top so you have a vertical asymptote at x=3
Let's check x=1
1^3-5(1)^2+4(1)-25=-25 so we have a vertical asymptote at x=1 also
There is no horizontal asymptote because degree of top is bigger than degree of bottom
There is a slant asympote because the degree of top is one more than degree of bottom (We can find this by doing long division)
x -1
--------------------------------------------------
x^2-4x+3 | x^3-5x^2+4x-25
- ( x^3-4x^2+3x)
--------------------------------
-x^2 +x -25
- (-x^2+4x-3)
---------------------
-3x-22
So the slant asymptote is to x-1
