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2 years ago

4. These polygons are similar, but not drawn to scale. Find the value of x.

Mathematics

1 answer:

Dafna11 [192]2 years ago

4 0

Answer

x=10

Step by step explanation:

It is a dilation to you would divide the known sides with each other.

8/2=4

16/4=4

So, you would then multiply 2.5 by 4 which would get you 10

preetty sure this is it, if its wrong then im sorry

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En una sala de teatro havian 7 filas 23 sillas cada una acaban de quitar 8 de las sillas ¿cuantas sillas quedan?

LenaWriter [7]

Answer:

Quedan 153 sillas

Step-by-step explanation:

Primero sabemos que en una sala de teatro tenían 7 filas, y en cada una de esas filas habían 23 sillas.

Entonces, originalmente, el número total de sillas es 7 veces 23:

S = 7*23 = 161

Luego sabemos que quitan 8 de esas sillas, entonces el número de sillas que quedan es la diferencia entre el número original de sillas, 161, y 8, que nos da:

S = 161 - 8 = 153

Quedan 153 sillas

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2 years ago

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tankabanditka [31]

Answer:

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3 years ago

Read 2 more answers

A jumbo crayon is composed of a cylinder with a conical tip. The cylinder is 12 cm tall with a radius of 1.5 cm, and the cone ha

s2008m [1.1K]

Answer:

Part 1) The lateral area of the cone is $LA=2\pi\ cm^{2}$

Part 2) The lateral surface area of the cylinder is $LA=36\pi\ cm^{2}$

Part 3) The surface area of the crayon is $SA=41.50\pi\ cm^{2}$

Step-by-step explanation:

Part 1) Find the lateral area of the cone

The lateral area of the cone is equal to

$LA=\pi rl$

we have

$r=1\ cm$

$l=2\ cm$

substitute

$LA=\pi (1)(2)$

$LA=2\pi\ cm^{2}$

Part 2) Find the lateral surface area of the cylinder

The lateral area of the cylinder is equal to

$LA=2\pi rh$

we have

$r=1.5\ cm$

$h=12\ cm$

substitute

$LA=2\pi (1.5)(12)$

$LA=36\pi\ cm^{2}$

Part 3) Find the surface area of the crayon

The surface area of the crayon is equal to the lateral area of the cone, plus the lateral area of the cylinder, plus the top area of the cylinder plus the bottom base of the crayon

Find the area of the bottom base of the crayon

$A=\pi[r2^{2}-r1^{2}]$

where

r2 is the radius of the cylinder

r1 is the radius of the cone

substitute

$A=\pi[1.5^{2}-1^{2}]$

$A=1.25\pi\ cm^{2}$

Find the area of the top base of the cylinder

$A=\pi(1.5)^{2}=2.25\pi\ cm^{2}$

Find the surface area

$SA=2\pi+36\pi+2.25\pi+1.25\pi=41.50\pi\ cm^{2}$

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3 years ago

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DaniilM [7]

Answer:

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Step-by-step explanation:

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3 years ago

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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

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

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2 years ago