Pls help me it’s due today

Mathematics

1 answer:

Marina86 [1]2 years ago

5 0

Answer:

x = 6.4

Step-by-step explanation:

Set up using proportion and cross multiply to solve:

8/10 = x/8

x = 6.4

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A rectangle with a length of L and a width of W has a diagonal of 10 inches. Express the perimeter P of the rectangle as a funct

KatRina [158]

<h2>Answer:</h2>

The expression which represents the perimeter P of the rectangle as a function of L is:

$Perimeter=2(L+\sqrt{100-L^2})$

<h2>Step-by-step explanation:</h2>

The length and width of a rectangle are denoted by L and W respectively.

Also the diagonal of a rectangle is: 10 inches.

We know that the diagonal of a rectangle in terms of L and W are given by:

$10=\sqrt{L^2+W^2}$

( Since, the diagonal of a rectangle act as a hypotenuse of the right angled triangle and we use the Pythagorean Theorem )

Hence, we have:

$10^2=L^2+W^2\\\\i.e.\\\\W^2=100-L^2\\\\W=\pm \sqrt{100-L^2}$

But we know that width can't be negative. It has to be greater than 0.

Hence, we have:

$W=\sqrt{100-L^2}$

Now, we know that the Perimeter of a rectangle is given by:

$Perimeter=2(L+W)$

Here we have:

$Perimeter=2(L+\sqrt{100-L^2})$

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

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