<span>625(5xy)^-3/ (5x)^2 y^7
625
= -------------------- / 25x^2y^7
125 x^3y^3
= 5/x^3y^3 / </span>25x^2y^7
= 5/x^3y^3 * (1/ 25x^2y^7)
= 1 / 5x^5y^10
answer
1
-----------------
5x^5y^10
Answer: 460.6 miles South of Chicago
Explanation:
1) <span>The City of New Orleans leaves Union Station in Chicago at 6:00 am and
travels south to New Orleans. The train travels the 930.0 miles to New Orleans at an average speed of 79.0 miles per hour.
Scheme:
→to south ←to north
|----------------------|---------------------|
A x X 930.0 - x B
6:00 am 6:00 am
→ ←
v = 79.0 mi/h v = 81.1 mi/h
=> x = 79.0 mi/h * t
2) at the same time the City of Chicago leaves New Orleans and travels north to Nhicago. The
train travels the 930.0 miles to Chicago at an average speed of 81.1
miles per hour.
930.0 - x = 81.1 mi/h * t
3) How many miles south of chicago do the two trains pass
each other?
Divide the two equations:
930.0 - x 81.1*t
------------- = ----------
x 79*t
t cancels out =>
930.0 - x 81.1
------------- = --------
x 79.0
Solve:
81.1x = 79.0(930.0 - x)
81.1x = 79.0 * 930.0 - 79.0x
81.1x + 79.0x = 73,470
=> 160.1x = 73,470
=> x = 73,470 / 160.1
=> x = 460.6 mi
4) Answer 460.6 miles South of Chicago.
</span>
Answer:
Step-by-step explanation:
Given is a differential equation as
Divide this by t to get in linear form
This is of the form
y' +p(t) y = Q(t)
where p(t) = 1/t
So solution would be
siubstitute y(1) = 16
Answer:
If A(t) represents the amount of salt in the tank at time t, the correct differential equation for A is is dA/dt = 15 - 0.005A
Option C) dA/dt = 15 - 0.005A is the correction Answer
Step-by-step explanation:
Given the data in the question;
If A(t) represents the amount of salt in the tank at time t, the correct differential equation for A is?
dA/dt = rate in - rate out
first we determine the rate in and rate out;
rate in = 3pound/gallon × 5gallons/min = 15 pound/min
rate out = A pounds/1000gallons × 5gallons/min = 5Ag/1000pounds/min
= 0.005A pounds/min
so we substitute
dA/dt = rate in - rate out
dA/dt = 15 - 0.005A
Therefore, If A(t) represents the amount of salt in the tank at time t, the correct differential equation for A is is dA/dt = 15 - 0.005A
Option C) dA/dt = 15 - 0.005A is the correction Answer
