They will have to bring in more than $600,000 a month to beat their competitors.
Step-by-step explanation:
Step 1; This establishment's competitors bring in $1,800,000 per quarter. This means that they bring in that amount of money through sales in a quarter of a year.
A quarter of a year =
× 12 months = 3 months.
So the competition brings in $1,800,000 in 3 months.
Step 2; Now we calculate how much this establishment must make to beat them.
Money to brought in a month = $1,800,000 / 3= $600,000 a month. So the team must bring in more than $600,000 a month to beat their competitor's sales of $1,800,000 in a quarter.
Answer:
y2 = C1xe^(4x)
Step-by-step explanation:
Given that y1 = e^(4x) is a solution to the differential equation
y'' - 8y' + 16y = 0
We want to find the second solution y2 of the equation using the method of reduction of order.
Let
y2 = uy1
Because y2 is a solution to the differential equation, it satisfies
y2'' - 8y2' + 16y2 = 0
y2 = ue^(4x)
y2' = u'e^(4x) + 4ue^(4x)
y2'' = u''e^(4x) + 4u'e^(4x) + 4u'e^(4x) + 16ue^(4x)
= u''e^(4x) + 8u'e^(4x) + 16ue^(4x)
Using these,
y2'' - 8y2' + 16y2 =
[u''e^(4x) + 8u'e^(4x) + 16ue^(4x)] - 8[u'e^(4x) + 4ue^(4x)] + 16ue^(4x) = 0
u''e^(4x) = 0
Let w = u', then w' = u''
w'e^(4x) = 0
w' = 0
Integrating this, we have
w = C1
But w = u'
u' = C1
Integrating again, we have
u = C1x
But y2 = ue^(4x)
y2 = C1xe^(4x)
And this is the second solution
Answer:
h(x) = |x+10|---------------------------------------------------------
Explanation:
To shift the graph 10 units to the left, we replace x with x+10. What's really going on is that the xy axis shifts 10 units to the right (because x is now x+10; eg, x = 2 ---> x+10 = 2+10 = 12) so it appears that the graph is moving to the left. The general rule is h(x) = g(x+10).
So,
g(x) = |x|
g(x+10) = |x+10| ... every x has been replaced with x+10
h(x) = g(x+10)
h(x) = |x+10|
We can use a graphing tool like GeoGebra to visually confirm we have the right answer (see attached). Note how a point like (0,0) on the green graph moves to (-10,0) on the red graph.