The answer would be disagree because it you reflected it over the x axis then it would be in the third quadrant
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Answer:
A
Step-by-step explanation:
![\left[\begin{array}{ccc}12 &-6\\1 &-10 \\\end{array}\right] +\left[\begin{array}{ccc}-2&14\\5&15\\\end{array}\right] =\left[\begin{array}{ccc}12-2&-6+14\\1+5&-10+15\\\end{array}\right] =\left[\begin{array}{ccc}10&8\\6&5\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D12%20%26-6%5C%5C1%20%26-10%20%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%2614%5C%5C5%2615%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D12-2%26-6%2B14%5C%5C1%2B5%26-10%2B15%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D10%268%5C%5C6%265%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Answer:
about 2283
Step-by-step explanation:
Plug 24 in for the value of t. e is its own value like pi but 2.1.........
Given 3a/(a+1)^2
To make the denominator a cube, you would have to multiply by 1 or (a+1)/(a+1)
yielding 3a(a+1)/(a+1)^3
(3a^2 + 3a) is the equivalent numerator
