Answer:
t_o = 3, so solution exists on (0,4).
Step-by-step explanation:
Use Theorem
Divide equation with t(t — 4).
y''+[3/(t-4)]*y'+ [4/t(t-4)]*y=2/t(t-4)
p(t)=3/t-4—> continuous on (-∞, 4) and (4,∞)
q(t) = 4/t(t-4) —> continuous on (-∞,0), (0,4) and (4, ∞)
g(t) = 2/t(t-4)—> continuous on (-∞, 0), (0,4) and (4,∞)
t_o = 3, so solution exists on (0,4).
Answer:
Step-by-step explanation:
<u>Number of those drive to school:</u>
<u>Number of those don't drive:</u>
<u>Probability:</u>
- P(Drive to school ' ) = 60/100 = 3/5
Answer:

Step-by-step explanation:

C: none of these are solutions to the given equation.
• If<em> y(x)</em> = <em>e</em>², then <em>y</em> is constant and <em>y'</em> = 0. Then <em>y'</em> - <em>y</em> = -<em>e</em>² ≠ 0.
• If <em>y(x)</em> = <em>x</em>, then <em>y'</em> = 1, but <em>y'</em> - <em>y</em> = 1 - <em>x</em> ≠ 0.
The actual solution is easy to find, since this equation is separable.
<em>y'</em> - <em>y</em> = 0
d<em>y</em>/d<em>x</em> = <em>y</em>
d<em>y</em>/<em>y</em> = d<em>x</em>
∫ d<em>y</em>/<em>y</em> = ∫ d<em>x</em>
ln|<em>y</em>| = <em>x</em> + <em>C</em>
<em>y</em> = exp(<em>x</em> + <em>C </em>)
<em>y</em> = <em>C</em> exp(<em>x</em>) = <em>C</em> <em>eˣ</em>
[tex]\begin{gathered} C=\text{ 72000 + 18x} \\ R=\text{ 27x } \\ \\ R\text{ }
Therefore, the company loses money if it sells less than 8000.
The answer is D.