Note that 2x-y=8 can be solved for y: y=2x-8. This is identical to the first equation. The graph of one is exactly the same as the graph of the other. Thus, the two equations are dependent.
im not sure this is right
g'(x) = 6b(-5x + 1)^5 (-5)
g'(x) = -30b(-5x +1)^5
g''(x) = -30b(5)(-5x + 1)^4 (-5)
g''(x) = 750b (-5x +1)^4
g(x) = b(−5x + 1)6 − a
when
g(-x) = b(5x +1)6 - a
g'(x) = -30b(-5x +1)^5 = 0
-5x +1 = 0
x = 15
Answer:
6.5 and 3.5
Step-by-step explanation:
Answer:
yes
Step-by-step explanation:
A proportional equation is of the form
y = kx where k is the constant of proportionality
y = 150x is proportional
Answer:
![V = \left[\begin{array}{ccc}5&-1\end{array}\right]](https://tex.z-dn.net/?f=V%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%26-1%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
We want to reflect this 2x1 vector on the line y = x.
To make this reflection we must use the following matrix:
![R=\left[\begin{array}{cc}0&1\\1&0\\\end{array}\right]](https://tex.z-dn.net/?f=R%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0%261%5C%5C1%260%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Where R is known as the reflection matrix on the line x = y
Now perform the product of the vector <-1,5> x R.
![\left[\begin{array}{ccc}-1\\5\end{array}\right]x\left[\begin{array}{ccc}0&1\\1&0\end{array}\right]\\\\\\\left[\begin{array}{ccc}-1(0) +5(1)&-1(1)+5(0)\end{array}\right]\\\\\\\left[\begin{array}{ccc}5&-1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%5C%5C5%5Cend%7Barray%7D%5Cright%5Dx%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%261%5C%5C1%260%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%280%29%20%2B5%281%29%26-1%281%29%2B5%280%29%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%26-1%5Cend%7Barray%7D%5Cright%5D)
The vector matrix that represents the reflection of the vector <-1,5> across the line x = y is:
