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3 years ago

HELP DUE IN 10 MINS! Use the Pythagorean Theorem to solve for x.

Mathematics

2 answers:

irakobra [83]3 years ago

8 0

Answer:

B. 5sqrt(2)

Step-by-step explanation:

x² = 4² + (√34)²
x² = 16 + 34
x² = 50
x = √50
x = 5√2

Elan Coil [88]3 years ago

5 0

$\huge\large \fbox \green{Answer :}$

B. 5sqrt(2)

$\huge\Large \fbox \green{Explanation:}$

We have

first side of triangle = √34

second side of triangle = 4

We need to find third side of triangle

let , the third side of triangle is x

( x)² = (√34 )² + (4)² ....( by Pythagoras Theorem)

x ² = 34 + 16

x ² = 50

x = √50

x = 5 √2

Hence, the third side is 5√2

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Step-by-step explanation:

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2 years ago

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

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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

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