Answer:
log9
Step-by-step explanation:
hello :
note : a>0 and b>0
1) log(ab) = loga+logb
2) loga^n = nloga.....n in N
log(9x^5) + 5 log (1/x) = log(9x^5) + log (1/x)^5 = log((9x^5) (1/x)^5)
= log(9x^5)/(x^5)) = log9 because : (1/x)^5 = 1/x^5
Answer:
You need to use the graph's equation and find the answer of y when x is 0 as this will give y when it is on the y axis.
To do this, substitute in 0 were x is in the equation and you will find the y intercept
Just put the coefients in to a matrix
1x-6y-3z=4
-2x+0y-3z=-8
-2x+2y-3z=-14
![\left[\begin{array}{ccc}1&-6&-3|4\\-2&0&-3|-8\\-2&2&-3|-14\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26-6%26-3%7C4%5C%5C-2%260%26-3%7C-8%5C%5C-2%262%26-3%7C-14%5Cend%7Barray%7D%5Cright%5D%20)
mulstiply 2nd row by -1 and add to 3rd
![\left[\begin{array}{ccc}1&-6&-3|4\\-2&0&-3|-8\\0&2&0|-6\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26-6%26-3%7C4%5C%5C-2%260%26-3%7C-8%5C%5C0%262%260%7C-6%5Cend%7Barray%7D%5Cright%5D)
divde last row by 2
![\left[\begin{array}{ccc}1&-6&-3|4\\-2&0&-3|-8\\0&1&0|-3\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26-6%26-3%7C4%5C%5C-2%260%26-3%7C-8%5C%5C0%261%260%7C-3%5Cend%7Barray%7D%5Cright%5D)
multiply 2rd row by 6 and add to top one
![\left[\begin{array}{ccc}1&0&-3|-14\\-2&0&-3|-8\\0&1&0|-3\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%26-3%7C-14%5C%5C-2%260%26-3%7C-8%5C%5C0%261%260%7C-3%5Cend%7Barray%7D%5Cright%5D)
multiply 1st row by -1 and add to 2nd
![\left[\begin{array}{ccc}1&0&-3|-14\\-3&0&0|6\\0&1&0|-3\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%26-3%7C-14%5C%5C-3%260%260%7C6%5C%5C0%261%260%7C-3%5Cend%7Barray%7D%5Cright%5D)
divide 2nd row by -3
![\left[\begin{array}{ccc}1&0&-3|-14\\1&0&0|-2\\0&1&0|-3\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%26-3%7C-14%5C%5C1%260%260%7C-2%5C%5C0%261%260%7C-3%5Cend%7Barray%7D%5Cright%5D)
mulstiply 2nd row by -1 and add to 1st row
![\left[\begin{array}{ccc}0&0&-3|-12\\1&0&0|-2\\0&1&0|-3\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%260%26-3%7C-12%5C%5C1%260%260%7C-2%5C%5C0%261%260%7C-3%5Cend%7Barray%7D%5Cright%5D)
divide 1st row by -3
![\left[\begin{array}{ccc}0&0&1|4\\1&0&0|-2\\0&1&0|-3\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%260%261%7C4%5C%5C1%260%260%7C-2%5C%5C0%261%260%7C-3%5Cend%7Barray%7D%5Cright%5D)
rerange
![\left[\begin{array}{ccc}1&0&0|-2\\0&1&0|-3\\0&0&1| 4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%260%7C-2%5C%5C0%261%260%7C-3%5C%5C0%260%261%7C%204%5Cend%7Barray%7D%5Cright%5D)
x=-2
y=-3
z=4
(x,y,z)
(-2,-3,4)
B is answer
Answer:
yeah you need i cleaner picture
Answer:

Step-by-step explanation:
This line was never finished, so you finish it like this:

3 = −⅖ + b

Parallel Lines have SIMILAR <em>RATE OF CHANGES</em> [<em>SLOPES</em>], so ⅖ remains the way it is.
I am joyous to assist you anytime.