Answer: Graph shifts 4 units to the left
Explanation:
I'm assuming you meant to say y = |x+4|
If so, then the graph shifts 4 units to the left. Replacing x with x+4 moves the xy axis 4 units to the right if we held the V shape in place (since each x is now 4 units larger). This gives the illusion the V shape is moving 4 units to the left.
Or you could look at the vertex point to see how it moves. On y = |x|, the vertex is at (0,0). It then moves to (-4,0) when we go to y = |x+4|
9514 1404 393
Answer:
see attached
Step-by-step explanation:
Polynomial long division is done the way any long division is done. Find a "partial quotient", subtract from the dividend the product of that partial quotient and the divisor. The result is a new dividend. Repeat until the degree of the dividend is less than that of the divisor.
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In the attached, the "Hints" show you how the partial quotient is found, and they show you how the product of the partial quotient and divisor is found.
The partial quotient term is simply the ratio of the highest degree terms of dividend and divisor. (Unlike numerical long division, there is no guessing.)
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The remainder is the dividend of lower degree than the divisor. As in numerical long division, the full quotient expresses the remainder over the divisor.
For example, 5 ÷ 3 = 1 r 2 = 1 + 2/3.
Your full quotient is (n+5) +1/(n-6).
True.
You're allowed to move stuff around. Dogs of the AMS.

We wrote it in the form dy/dx = g(x) f(y) so this is separable.
Answer:
y = 2x + 1 --> linear
y = -4x + 7 --> non-linear
Not a solution for linear system.
Step-by-step explanation:
for (a), y = 2x+1, substitute the x and y values. keep in mind, that in a linear pair, (x, y). So, for the first equation you get:
7 = 2x3 + 1. This is correct, because 6 + 1 is 7. Therefore, (a) is linear.
for (b), we have to substitute our values again. You get:
7 = -4x3 + 7, which is
7 = -12+7, which is not true. So, (b) is not linear.
This means that for the linear pair (3, 7), it does not satisfy both equations, which means that it is not a solution for the linear system.