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3 years ago

Can someone tell me how to do this? with steps?

Mathematics

1 answer:

MrRa [10]3 years ago

3 0

Answer:

Step-by-step explanation:

In Δ AFB,

∠AFB + ∠ABF + ∠A = 180 {Angle sum property of triangle}

90 + 48 + ∠1 = 180

138 + ∠1 = 180

∠1 = 180 - 138

∠1 = 42°

FC // ED and FD is transversal

So, ∠CFD ≅∠EDF {Alternate interior angles are congruent}

∠2 = 39°

In ΔFCD,

∠2 + ∠3 + ∠FCD = 180

39 + ∠3 + 90 = 180

129 +∠3 = 180

∠3 = 180- 129

∠3 = 51°

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As his empire spread throughout the ancient world, Alexander founded roughly how many cities?

Studentka2010 [4]

About 70 Cities in Total.

For his more noteworthy grandness, Alexander established exactly 70 urban areas in the grounds he vanquished and requested them named after himself. Most celebrated, obviously, is Alexandria in Egypt. In India, when his dearest horse kicked the bucket, he requested a city to be constructed named Bucephala.

6 0

4 years ago

Find the percentage of vacationers from<br> question 10 who spent between $1500<br> and $2000.

mars1129 [50]

Answer:

The percentage change of vacationers is 33.3 %

Step-by-step explanation:

Given as :

The old value of the vacationers = $ 1500

The new value of the vacationers = $ 2000

Let the percentage variation = x %

Or, x % increase = $\dfrac{\tyextrm new value - \textrm old value}{\textrm old value}$ × 100

Or, x % increase = $\dfrac{\tyextrm $ 2000 - \textrm $ 1500}{\textrm $ 1500}$ × 100

Or, x % increase = $\frac{500}{1500}$ × 100

Or, x % increase = $\frac{1}{3}$ × 100

Or, x % increase = $\frac{100}{3}$

∴ x = 33.3 %

So, The percentage increase change is 33.3 %

Hence The percentage change of vacationers is 33.3 % Answer

3 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

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