How many are 15 x 15 ?

Mathematics

2 answers:

andrew11 [14]3 years ago

7 0

Answer:

225

Step-by-step explanation:

RSB [31]3 years ago

7 0

Answer:

225

Step-by-step explanation:

x15

/225

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$y=-5x+7$

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3 years ago

What is the solution to the equation?

ki77a [65]

Answer:

Step-by-step explanation:

$3 {(x + 9)}^{ \frac{3}{4} } = 24 \\ \\ {(x + 9)}^{ \frac{3}{4} } = \frac{24}{3} \\ \\ {(x + 9)}^{ \frac{3}{4} } = 8 \\ \\ {(x + 9)}^{ \frac{3}{4} } = {2}^{3} \\ \\ (x + 9) = {2}^{3 \times \frac{4}{3} } \\ \\ x + 9 = {2}^{4} \\ \\ x + 9 = 16 \\ \\ x = 16 - 9 \\ \\ x = 7$

7 0

3 years ago

D^2(y)/(dx^2)-16*k*y=9.6e^(4x) + 30e^x

MA_775_DIABLO [31]

The solution depends on the value of $k$ . To make things simple, assume $k>0$ . The homogeneous part of the equation is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}-16ky=0$

and has characteristic equation

$r^2-16k=0\implies r=\pm4\sqrt k$

which admits the characteristic solution $y_c=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}$ .

For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be $y_p=ae^{4x}+be^x$ . Then

$\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=16ae^{4x}+be^x$

So you have

$16ae^{4x}+be^x-16k(ae^{4x}+be^x)=9.6e^{4x}+30e^x$
$(16a-16ka)e^{4x}+(b-16kb)e^x=9.6e^{4x}+30e^x$

This means

$16a(1-k)=9.6\implies a=\dfrac3{5(1-k)}$
$b(1-16k)=30\implies b=\dfrac{30}{1-16k}$

and so the general solution would be

$y=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}+\dfrac3{5(1-k)}e^{4x}+\dfrac{30}{1-16k}e^x$