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zvonat [6]
3 years ago
15

PLEASE HELP Find the length of the missing side.

Mathematics
1 answer:
BabaBlast [244]3 years ago
3 0

Answer:

use Pythagoras theorem

the missing side is base

Step-by-step explanation:

pythagoras theorem

h^2 = p^2 + b^2

73^2 = 48^2 + b^2

5329 = 2304 + b^2

5329 - 2304 = b^2

3025 = b^2

√3025 = b

55 = b

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Use the substitution of x=e^{t} to transform the given Cauchy-Euler differential equation to a differential equation with consta
kherson [118]

By the chain rule,

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}\implies\dfrac{\mathrm dy}{\mathrm dt}=x\dfrac{\mathrm dy}{\mathrm dx}

which follows from x=e^t\implies t=\ln x\implies\dfrac{\mathrm dt}{\mathrm dx}=\dfrac1x.

\dfrac{\mathrm dy}{\mathrm dt} is then a function of x; denote this function by f(x). Then by the product rule,

\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1x\dfrac{\mathrm dy}{\mathrm dt}\right]=-\dfrac1{x^2}\dfrac{\mathrm dy}{\mathrm dt}+\dfrac1x\dfrac{\mathrm df}{\mathrm dx}

and by the chain rule,

\dfrac{\mathrm df}{\mathrm dx}=\dfrac{\mathrm df}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}=\dfrac1x\dfrac{\mathrm d^2y}{\mathrm dt^2}

so that

\dfrac{\mathrm d^2y}{\mathrm dt^2}-\dfrac{\mathrm dy}{\mathrm dt}=x^2\dfrac{\mathrm d^2y}{\mathrm dx^2}

Then the ODE in terms of t is

\dfrac{\mathrm d^2y}{\mathrm dt^2}+8\dfrac{\mathrm dy}{\mathrm dt}-20y=0

The characteristic equation

r^2+8r-20=(r+10)(r-2)=0

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y_c(t)=C_1e^{-10t}+C_2e^{2t}

Solving in terms of x gives

y_c(x)=C_1e^{-10\ln x}+C_2e^{2\ln x}\implies\boxed{y_c(x)=C_1x^{-10}+C_2x^2}

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3 years ago
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