Answer:
Hence total number freshman presented are 655.
The system is relative system as there is correlation between freshman and
sophomores<em>.</em><em>( as freshman are 60 more than the Sophomores)</em>
Step-by-step explanation:
Given:
1250 total students of sophomores and freshman.
To Find:
freshman count and its type of system used.
Solution:
We have that ,
consider x be the no.of freshman and y be the sophomores
So by given condition,
x+y=1250 ,..............................Equation(1)
And other one,
x=60+y
Use above value in equation (1) we get ,
60+y+y=1250
2y=1250-60
y=595
Now number of freshman ,
x+y=1250
x=1250-595
x=655
Hence total number freshman presented are 655.
The system represent the relative proportion system.The break even and total value should include all students in university .
Relative means in relationship with one another .
As there are 60 more number of freshman than the sophomores.
Answer:
yes it is
Step-by-step explanation:
The answer would be 11 gallons.
The tub starts with 32 gallons. Every minute it loses 3 gallons.
So after 1 minute it has lost 3 gallons. For example:
32-3= 29 Meaning that after 1 minute the tub has 29 gallons left.
Now you have to remember that the tub is draining for 7 minutes. So it is losing 3 gallons 7 times, because it loses 3 gallons each minute and there are 7 minutes.
We can use multiplication to find how many gallons the tub loses after 7 minutes. This sign “X” basically means “groups of”. We have 7 groups of 3, or
7 X 3
This is the same as saying we have 3, 7 times. Written like this:
3+3 +3 +3 +3 +3 +3 = 21
So after 7 minutes the tub has lost 21 gallons.
Now we take the original number of gallons and take away what was lost:
32-21=11
So there are 11 gallons left after 7 minutes.
Please let me know if you need any further explanation. Hope this helped.
Answer:

Step-by-step explanation:
well, by using definition of tan of an angle
let the distance between Earth and shooting star is
.
since
,
then 
The corresponding homogeneous ODE has characteristic equation
with roots at
, thus admitting the characteristic solution

For the particular solution, assume one of the form



Substituting into the ODE gives



Then the general solution to this ODE is



Assume a solution of the form



Substituting into the ODE gives



so the solution is



Assume a solution of the form


Substituting into the ODE gives



so the solution is
