What is the answer to solvefor A,y=Ax^2+Be^x ?

The answer to solvefor A,y=Ax^2+Be^x is A=(y-Be^x)/(x^2);\quad x\ne 0

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solvefor A,y=Ax^2+Be^x

solve for

A, y = A x^{2} + B e^{x}

A = \frac{y - B e ^{x}}{x ^{2}}; x \neq = 0

Solution steps

y = A x^{2} + B e^{x}

Switch sides

A x^{2} + B e^{x} = y

Move

B e^{x}

to the right side

A x^{2} = y - B e^{x}

Divide both sides by

x^{2}; x \neq = 0

A = \frac{y - B e ^{x}}{x ^{2}}; x \neq = 0

f = lo g_{\frac{1}{4}} (14 + 2)

4 (x - 2) = - 8 (x - (- 4))

y = m (x - x^{1}) + y^{1}

x, - 7 x - 2 y = - 21

8 = 4 (q - 2) + 4

What is the answer to solvefor A,y=Ax^2+Be^x ?
The answer to solvefor A,y=Ax^2+Be^x is A=(y-Be^x)/(x^2);\quad x\ne 0