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geniusboy [140]

1 year ago

12

Tim wants to build a rectangular fence around his yard. He has 42 feet of fencing. If he wants the length to be twice the width.

What is the largest possible length

1 answer:

gtnhenbr [62]1 year ago

8 0

Answer: The largest he can possibly build is x = 14

Step-by-step explanation:

You might be interested in

Factor out the greatest common factor. <br>2f^3-8f^2

ella [17]

Answer:

Step-by-step explanation:

8f² = 2³f²

GCF of 2f³ and 8f² = 2f²

2f³ - 8f² = 2f²(f - 4f)

8 0

3 years ago

Find a particular solution to <img src="https://tex.z-dn.net/?f=%20x%5E%7B2%7D%20%20%5Cfrac%7B%20d%5E%7B2%7Dy%20%7D%7Bd%20x%5E%7

Digiron [165]

$y=x^r$
$\implies r(r-1)x^r+6rx^r+4x^r=0$
$\implies r^2+5r+4=(r+1)(r+4)=0$
$\implies r=-1,r=-4$

so the characteristic solution is

$y_c=\dfrac{C_1}x+\dfrac{C_2}{x^4}$

As a guess for the particular solution, let's back up a bit. The reason the choice of $y=x^r$ works for the characteristic solution is that, in the background, we're employing the substitution $t=\ln x$ , so that $y(x)$ is getting replaced with a new function $z(t)$ . Differentiating yields

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac1x\dfrac{\mathrm dz}{\mathrm dt}$
$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac1{x^2}\left(\dfrac{\mathrm d^2z}{\mathrm dt^2}-\dfrac{\mathrm dz}{\mathrm dt}\right)$

Now the ODE in terms of $t$ is linear with constant coefficients, since the coefficients $x^2$ and $x$ will cancel, resulting in the ODE

$\dfrac{\mathrm d^2z}{\mathrm dt^2}+5\dfrac{\mathrm dz}{\mathrm dt}+4z=e^{2t}\sin e^t$

Of coursesin, the characteristic equation will be $r^2+6r+4=0$ , which leads to solutions $C_1e^{-t}+C_2e^{-4t}=C_1x^{-1}+C_2x^{-4}$ , as before.

Now that we have two linearly independent solutions, we can easily find more via variation of parameters. If $z_1,z_2$ are the solutions to the characteristic equation of the ODE in terms of $z$ , then we can find another of the form $z_p=u_1z_1+u_2z_2$ where

$u_1=-\displaystyle\int\frac{z_2e^{2t}\sin e^t}{W(z_1,z_2)}\,\mathrm dt$
$u_2=\displaystyle\int\frac{z_1e^{2t}\sin e^t}{W(z_1,z_2)}\,\mathrm dt$

where $W(z_1,z_2)$ is the Wronskian of the two characteristic solutions. We have

$u_1=-\displaystyle\int\frac{e^{-2t}\sin e^t}{-3e^{-5t}}\,\mathrm dt$
$u_1=\dfrac23(1-2e^{2t})\cos e^t+\dfrac23e^t\sin e^t$

$u_2=\displaystyle\int\frac{e^t\sin e^t}{-3e^{-5t}}\,\mathrm dt$
$u_2=\dfrac13(120-20e^{2t}+e^{4t})e^t\cos e^t-\dfrac13(120-60e^{2t}+5e^{4t})\sin e^t$

$\implies z_p=u_1z_1+u_2z_2$
$\implies z_p=(40e^{-4t}-6)e^{-t}\cos e^t-(1-20e^{-2t}+40e^{-4t})\sin e^t$

and recalling that $t=\ln x\iff e^t=x$ , we have

$\implies y_p=\left(\dfrac{40}{x^3}-\dfrac6x\right)\cos x-\left(1-\dfrac{20}{x^2}+\dfrac{40}{x^4}\right)\sin x$

4 0

2 years ago

Solve the system of equations given below.

Yuri [45]

Answer:

C. $(-1, 10)$

Step-by-step explanation:

{5 = 10 - 5 ☑

{−20 = 1 - 21 ☑

This is what you will get when you plug these coordinates into the system of equations.

I am joyous to assist you anytime.

4 0

3 years ago

Please answer asap. there are two pics :)

Thepotemich [5.8K]

Answer:

$\boxed{\sf A. \ 0.34}$

Step-by-step explanation:

The first triangle is a right triangle and it has one acute angle of 70 degrees.

We can approximate $\sf \frac{WY}{WX}$ from right triangle 1.

The side adjacent to 70 degrees is WY. The side or hypotenuse is WX.

The side adjacent to 70 degrees in right triangle 1 is 3.4. The side or hypotenuse is 10.

$\sf \frac{3.4}{10} =0.34$

8 0

3 years ago

Solve for y -5-10=-60 please help

Ket [755]

Answer:

y=-45

Step-by-step explanation:

y-15=-60

y=-45

8 0

3 years ago

Read 2 more answers