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Zigmanuir [339]

3 years ago

11

Cesar comenta que su papa esta interesado en comprar un terreno o vivienda.En estos tiempos de cuarentena su padre ha aprendido

a comunicarse mas por medio de las redes sociales y justamente,atraves de whappsap,un amigo le envio una imagen del terreno de una vivienda en venta. Su padre le mostro la imagen y quiere que le alyude a realizar diversos calculos.si los calculos son 40m+14m+a+20m+54m+20+a.¿cual es el area de la vivienda?

1 answer:

quester [9]3 years ago

3 0

Answer:

El área de la casa es de 1,304 m²

Step-by-step explanation:

De las dimensiones dadas en línea, tenemos que el área del jardín = 640 m²

Las dimensiones del jardín son;

Longitud = 40 m.

Ancho = a m

Por lo tanto, 40 × a = 640

a = 640/40 = 16 m

La dimensión del compuesto es así;

Longitud = 40 + 14 = 54 m

Ancho = 20 + a = 20 + 16 = 36 m

El área del compuesto = Área de la casa + Área del jardín

El área del compuesto = Ancho del compuesto × Longitud del compuesto

El área del compuesto = 36 × 54 = 1,944 m²

∴ 1,944 m² = Área de la casa + Área del jardín

1,944 m² = Área de la casa + 640 m²

Área de la casa = 1,944 m² - 640 m² = 1,304 m²

El área de la casa = 1,304 m².

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Which is the inverse of the function f(x) = x2 − 4?

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The function is not one-to-one function, hence the inverse doesn't exist for the whole domain of the function. But if we restrict the domain to $x\in(-\infty,0]$ or $x\in[0,\infty)$ , then we can give the inverse.

$f(x)=x^2-4\\\\y=x^2-4\\\\x^2=y+4\\\\x=\sqrt{y+4} \vee x=-\sqrt{y+4}\\\\f^{-1}(x)=\sqrt{x+4}\vee f^{-1}(x)=-\sqrt{x+4}$

So, when $D_{f(x)}:x\in(-\infty,0]$ , the inverse is $f^{-1}(x)=-\sqrt{x+4}$ and when $D_{f(x)}:x\in[0,\infty)$ , the inverse is $f^{-1}(x)=\sqrt{x+4}$ .

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3 years ago

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3 years ago

Two particles move in the xy-plane. At time t, the position of particle A is given by x(t)=5t−5 and y(t)=2t−k, and the position

Ipatiy [6.2K]

Answer:

Part A) Not collide

Part B) k = 4

Part C) Particle B is moving fast.

Step-by-step explanation:

Two particles move in the xy-plane. At time, t

<u><em>Position of particle A:-</em></u>

$x(t)=5t-5$

$y(t)=2t-k$

<em><u>Position of particles B:-</u></em>

$x(t)=4t$

$y(t)=t^2-2t+1$

Part A) For k = -6

Position particle A, (5t-5,2t+6)

Position of particle B, $(4t,t^2-2t-1)$

If both collides then x and y coordinate must be same

Therefore,

For x-coordinate:

5t - 5 = 4t

t = 5

For y-coordinate:

$2t+6=t^2-2t-1$

$t^2-4t-7=0$

$t=-1.3,5.3$

The value of t is not same. So, k = -6 A and B will not collide.

Part B) If both collides then x and y coordinate must be same

For x-coordinate:

5t - 5 = 4t

t = 5

For y-coordinate:

$2t-k=t^2-2t-1$

Put t = 5

$10-k=25-10-1$

$k=4$

Hence, if k = 4 then A and B collide.

Part C)

Speed of particle A, $\dfrac{dA}{dt}$

$\dfrac{dA}{dt}=\dfrac{dy}{dt}\cdot \dfrac{dt}{dx}$

$\dfrac{dA}{dt}=2\cdot \dfrac{1}{5}\approx 0.4$

Speed of particle B, $\dfrac{dB}{dt}$

$\dfrac{dB}{dt}=\dfrac{dy}{dt}\cdot \dfrac{dt}{dx}$

$\dfrac{dB}{dt}=2t-2\cdot \dfrac{1}{4}$

At t = 5

$\dfrac{dB}{dt}=10-2\cdot \dfrac{1}{4}=2$

Hence, Particle B moves faster than particle A

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