Part A.
You need two equations with the same slope and different y-intercepts.
Their graph is parallel lines. Since the lines do not intersect, there is no solution.
y = 2x + 2
y = 2x - 2
Part B.
We use the first equation as above. For the second equation, we use an equation with different slope. Two lines with different slopes always intersect.
y = 2x + 2
y = -2x - 2
In the second equation, y = -2x - 2. We now substitute -2x - 2 for y in the first equation.
-2x - 2 = 2x + 2
-4x = 4
x = -1
Now substitute -1 for x in the first equation to find y.
y = 2x + 2
y = 2(-1) + 2
y = -2 + 2
y = 0
Solution: x = -1 and y = 0
Answer:
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Step-by-step explanation:
Remember, y=mx+b
y is equal to any given y point
x is equal to any given x point
m is equal to slope
b is equal to the y-intercept, or where x = 0 and the line crosses the horizon line.
In order to graph the line correctly, you have to isolate y.
y+x=-3
y=-x-3 would be equal to y=mx+b format
slope is negative 1
y intercept is negative 3
start on the y line, go to (0,-3) and start your line.
slope is negative 1, so you go down one and right one.
Answer: f(x) = 1^(x + 1)
Step-by-step explanation:
we have that h(x) = 1^x
and h(x) = f(g(x))
This mean that we are evaluating the function f(y) in the point y = g(x)
where g(x) = x - 1
then:
f(g(x) = f(x - 1) = h(x) = 1^x
then we should have that:
f(x) = 1^(x + 1)
then:
f(x - 1) = 1^(x - 1 + 1) = 1^x
A function has a vertical asymptote where it is not defined, and the logarithm is defined only if its argument is strictly positive: it has a vertical asymptote if its argument is zero, and is not defined if its argument is negative.
So, you have a vertical asymptote where
Answer:
Step-by-step explanation:
F = 
mv² = Fr
m = 
