Substitute 
, so that 
, and
which is separable as
Integrate both sides with respect to 
. For the integral on the left, first split into partial fractions:
Solve for 
:
Replace and solve for 
:
Now use the given initial condition to solve for 
:
so that the particular solution is
The answer is: [B]: " x + 3y + 10 = 0 " .
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Explanation:
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Note the equation for a line; in 'slope-intercept form': "y = mx + b" ;
in which "y" is isolated alone, as a single variable, on the left-hand side of the equation;
"m" = the slope of the line; and is the co-efficient of "x" ;
b = the "y-intercept"; (or the "y-coordinate of the point of the "y-intercept").
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So, given the information in this very question/problem:
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slope = m = (-1/3) ;
b = y-intercept = (10/3) ;
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And we can write the equation of the line; in "slope-intercept form"; that is:
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" y = mx + b " ; as:
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" y = (-1/3)x + (10/3) " ;
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Now, the problem asks for the equation of this line; in "general form"; or "standard format"; which is:
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"Ax + By + C = 0 " ;
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So; given:
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" y = (-1/3)x + (10/3) " ;
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We can multiply the ENTIRE EQUATION (both sides) by "3" ; to get rid of the "fractions" ;
→ 3* { y = (-1/3)x + (10/3) } ;
→ 3y = -1x + 10 ;
↔ -1x + 10 = 3y ;
Subtract "(3y)" from each side of the equation:
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-1x + 10 − 3y = 3y − 3y ;
to get:
-1x + 10 − 3y = 0 ;
↔ -1x − 3y − 10 = 0 ;
→ This is not one of the "3 (THREE) answer choices given" ;
→ So, multiply the ENTIRE EQUATION (both sides); by "-1" ; as follows:
-1 * {-1x − 3y − 10 = 0} ;
to get:
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→ " x + 3y + 10 = 0 " ; which is: "Answer choice: [B] ."
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Note that is equation is in the "standard format" ;
→ " Ax + By + C = 0 " .
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Answer:
The slope of y = 1 is 0
explanation:
Determining slope has been based on a simple formula, y2-y1 over x2-x1.
It is crucial to follow this formula in order to attain the proper slope.
When a simple expression is given, for instance; y = 1 then it could be divided by 0; thus, resulting in the answer simply being zero.
However, x=1 would contain the slope of undefined because any number divided by zero results in an undefined answer.
1.
If no changes are made, the school has a revenue of :
625*400$/student=250,000$
2.
Assume that the school decides to reduce n*20$.
This means that there will be an increase of 50n students.
Thus there are 625 + 50n students, each paying 400-20n dollars.
The revenue is:
(625 + 50n)*(400-20n)=12.5(50+n)*20(20-n)=250(n+50)(20-n)
3.
check the options that we have,
a fee of $380 means that n=1, thus
250(n+50)(20-n)=250(1+50)(20-1)=242,250 ($)
a fee of $320 means that n=4, thus
250(n+50)(20-n)=250(4+50)(20-4)=216,000 ($)
the other options cannot be considered since neither 400-275, nor 400-325 are multiples of 20.
Conclusion, neither of the possible choices should be applied, since they will reduce the revenue.
