URGENT PLEASE HELP!!

Mathematics

1 answer:

Lostsunrise [7]1 year ago

3 0

Answer:

the answer I got is a=-the sign and but 2b under it +7

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The base of a triangular prism is a right triangle with a base of 5 inches, a height of 12 inches, and a third side length of 13

myrzilka [38]

Answer:

690 sq in

Step-by-step explanation:

SA = LA + 2B where LA is the lateral area and B is the area of the base

The triangular base has an area of

B = 1/2bh 1/2(5)(12) = 30

LA = ph where p is the perimeter of the base and h is the height of the prism

LA = (12 + 5 + 13)(21) = 30(21) = 630

SA = 630 + 2(30) = 630 + 60 = 690 sq in

3 0

2 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

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