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lana [24]
3 years ago
7

Students have a grace period of ______ before they must begin repaying Direct Stafford Loans.

Mathematics
2 answers:
Nady [450]3 years ago
6 0
D. 6. months. This can be extended for a further three years, however.
Mnenie [13.5K]3 years ago
6 0

Answer: The answer is D) 6 months

APEX

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F(x) = 2x2 - 5<br> and<br> g(x) = 5x + 7<br> Find<br> f(g(x)) = [ ? ]x2 + [ ]x +
lbvjy [14]

Answer:10x^2 -18

Step-by-step explanation:since f(x) is becoming the new x then youd subtitute the equation into the x of the g(x) equation. which would be 5(2x^2 -5) +7 which is 10x^2 -25 +7.and you'd simplify from there.

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3 years ago
Consider the differential equation y'' − y' − 20y = 0. Verify that the functions e−4x and e5x form a fundamental set of solution
KIM [24]

Answer:

Therefore e^{-4x} and e^{5x} are fundamental solution of the given differential equation.

Therefore  e^{-4x} and e^{5x} are linearly independent, since W(e^{-4x},e^{5x})=9e^x\neq 0

The general solution of the differential equation is

y=c_1e^{-4x}+c_2e^{5x}

Step-by-step explanation:

Given differential equation is

y''-y'-20y =0

Here P(x)= -1, Q(x)= -20 and R(x)=0

Let trial solution be y=e^{mx}

Then, y'=me^{mx}   and   y''=m^2e^{mx}

\therefore m^2e^{mx}-m e^{mx}-20e^{mx}=0

\Rightarrow m^2-m-20=0

\Rightarrow m^2-5m+4m-20=0

\Rightarrow m(m-5)+4(m-5)=0

\Rightarrow (m-5)(m+4)=0

\Rightarrow m=-4,5

Therefore the complementary function is = c_1e^{-4x}+c_2e^{5x}

Therefore e^{-4x} and e^{5x} are fundamental solution of the given differential equation.

If y_1 and y_2 are the fundamental solution of differential equation, then

W(y_1,y_2)=\left|\begin{array}{cc}y_1&y_2\\y'_1&y'_2\end{array}\right|\neq 0

Then  y_1 and y_2 are linearly independent.

W(e^{-4x},e^{5x})=\left|\begin{array}{cc}e^{-4x}&e^{5x}\\-4e^{-4x}&5e^{5x}\end{array}\right|

                    =e^{-4x}.5e^{5x}-e^{5x}.(-4e^{-4x})

                    =5e^x+4e^x

                   =9e^x\neq 0

Therefore  e^{-4x} and e^{5x} are linearly independent, since W(e^{-4x},e^{5x})=9e^x\neq 0

Let the the particular solution of the differential equation is

y_p=v_1e^{-4x}+v_2e^{5x}

\therefore v_1=\int \frac{-y_2R(x)}{W(y_1,y_2)} dx

and

\therefore v_2=\int \frac{y_1R(x)}{W(y_1,y_2)} dx

Here y_1= e^{-4x}, y_2=e^{5x},W(e^{-4x},e^{5x})=9e^x ,and  R(x)=0

\therefore v_1=\int \frac{-e^{5x}.0}{9e^x}dx

       =0

and

\therefore v_2=\int \frac{e^{5x}.0}{9e^x}dx

       =0

The the P.I = 0

The general solution of the differential equation is

y=c_1e^{-4x}+c_2e^{5x}

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