Answer: The first one is not a function the second is a function the third one is not one and the fourth one is
Step-by-step explanation:
Answer:
The product is:
![\left[\begin{array}{cc}15 & 14\\-1 & 9\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D15%20%26%2014%5C%5C-1%20%26%209%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
For this problem you need to multiply the first row only for the two first column of the others matrix and get the desired result:
![\left[\begin{array}{ccc}1&3&1\\-2&1&0\end{array}\right] \times \left[\begin{array}{cc}2&-2\\3&5\\4&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%263%261%5C%5C-2%261%260%5Cend%7Barray%7D%5Cright%5D%20%5Ctimes%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%26-2%5C%5C3%265%5C%5C4%261%5Cend%7Barray%7D%5Cright%5D)

So the value of the element in the position
is 15

So the value of the element in the position
is 14
Then with these two values you can determinate the result matrix.
![\left[\begin{array}{cc}15 & 14\\-1 & 9\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D15%20%26%2014%5C%5C-1%20%26%209%5Cend%7Barray%7D%5Cright%5D)
Answer:
50
Step-by-step explanation:
2(3)^2+9(3)+5
2(9)+27+5
18+27+5
=50


- <u>A </u><u>triangle </u><u>with </u><u>sides </u><u>11m</u><u>, </u><u> </u><u>13m </u><u>and </u><u>18m</u>

- <u>We</u><u> </u><u>have </u><u>to </u><u>check </u><u>it </u><u>whether </u><u>it </u><u>is </u><u>right </u><u>angled </u><u>triangle </u><u>or </u><u>not</u><u>? </u>


According to the Pythagoras theorem, The sum of the squares of perpendicular height and the square of the base of the triangle is equal to the square of hypotenuse that is sum of the squares of two small sides equal to the square of longest side of the triangle.
<u>We </u><u>imply</u><u> </u><u>it </u><u>in </u><u>the </u><u>given </u><u>triangle </u><u>,</u>





<u>From </u><u>Above </u><u>we </u><u>can </u><u>conclude </u><u>that</u><u>, </u>
The sum of the squares of two small sides that is perpendicular height and base is not equal to the square of longest side that is Hypotenuse

B I think hopes this helps