Answer:
12x+18/x+5
Step-by-step explanation:
x = -2
function value at x = -2 is intersection of the x value and graph
which is, 3
Hope it is clear:)
C: none of these are solutions to the given equation.
• If<em> y(x)</em> = <em>e</em>², then <em>y</em> is constant and <em>y'</em> = 0. Then <em>y'</em> - <em>y</em> = -<em>e</em>² ≠ 0.
• If <em>y(x)</em> = <em>x</em>, then <em>y'</em> = 1, but <em>y'</em> - <em>y</em> = 1 - <em>x</em> ≠ 0.
The actual solution is easy to find, since this equation is separable.
<em>y'</em> - <em>y</em> = 0
d<em>y</em>/d<em>x</em> = <em>y</em>
d<em>y</em>/<em>y</em> = d<em>x</em>
∫ d<em>y</em>/<em>y</em> = ∫ d<em>x</em>
ln|<em>y</em>| = <em>x</em> + <em>C</em>
<em>y</em> = exp(<em>x</em> + <em>C </em>)
<em>y</em> = <em>C</em> exp(<em>x</em>) = <em>C</em> <em>eˣ</em>
Answer:
x= -18
Step-by-step explanation:
Solve for
x
by simplifying both sides of the equation, then isolating the variable.
Answer:
x = 1
Step-by-step explanation:
Given the matrix;
![-2\left[\begin{array}{ccc}x&-1\\3&5\end{array}\right] + \left[\begin{array}{ccc}3&8\\-1&6\end{array}\right] = \left[\begin{array}{ccc}x&10\\-7&-4\end{array}\right]](https://tex.z-dn.net/?f=-2%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%26-1%5C%5C3%265%5Cend%7Barray%7D%5Cright%5D%20%2B%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%268%5C%5C-1%266%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%2610%5C%5C-7%26-4%5Cend%7Barray%7D%5Cright%5D)
Converting to equations
From the first row of each matrices
-2x + 3 = x
Collect the like terms
-2x - x = -3
-3x = -3
x = -3/-3
x = 1
Hence the value of x is 1
