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STALIN [3.7K]
3 years ago
9

Hola, estoy empezando a ver trigonometria queria ver si me pueden ayudar con este problema: Un extraterrestre diminuto pero horr

ible está parado en la punta del edificio Empire State (que mide 443443443 metros de alto) y ¡amenaza con destruir la ciudad de Nueva York! Un agente de los Hombres de Negro está parado al nivel del piso, a 181818 metros del edificio, y apunta su pistola láser al extraterrestre. A qué ángulo, en grados, debe el agente disparar su pistola láser? Redondea tu respuesta final a la décima más cercana.
Mathematics
1 answer:
Kryger [21]3 years ago
3 0

Answer:

El agente debe disparar a un ángulo aproximado de 87.7° con respecto a la horizontal.

Step-by-step explanation:

Sea E la localización del extraterrestre, O, el del hombre de negro, y B, el de la base del edificio. A partir de la información disponible, construimos el diagrama geométrico que indica la presencia de un triángulo rectángulo, el cual puede consultarse en la imagen adjunta.

Podemos determinar el ángulo de disparo mediante la siguiente relación trigonométrica inversa:

A = \tan^{-1} \frac{443\,m}{18\,m}

A \approx 87.7^{\circ}

El agente debe disparar a un ángulo aproximado de 87.7° con respecto a la horizontal.

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Find the length of side a.<br> 5<br> 3<br> O A. 16<br> O B. 34<br> O c. 2<br> O D. 4
xxMikexx [17]

Answer:

D

Step-by-step explanation:

a=\sqrt{c^{2} -b^{2} }

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5 0
3 years ago
Solve y'' + 10y' + 25y = 0, y(0) = -2, y'(0) = 11 y(t) = Preview
svetlana [45]

Answer:  The required solution is

y=(-2+t)e^{-5t}.

Step-by-step explanation:   We are given to solve the following differential equation :

y^{\prime\prime}+10y^\prime+25y=0,~~~~~~~y(0)=-2,~~y^\prime(0)=11~~~~~~~~~~~~~~~~~~~~~~~~(i)

Let us consider that

y=e^{mt} be an auxiliary solution of equation (i).

Then, we have

y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.

Substituting these values in equation (i), we get

m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.

So, the general solution of the given equation is

y(t)=(A+Bt)e^{-5t}.

Differentiating with respect to t, we get

y^\prime(t)=-5e^{-5t}(A+Bt)+Be^{-5t}.

According to the given conditions, we have

y(0)=-2\\\\\Rightarrow A=-2

and

y^\prime(0)=11\\\\\Rightarrow -5(A+B\times0)+B=11\\\\\Rightarrow -5A+B=11\\\\\Rightarrow (-5)\times(-2)+B=11\\\\\Rightarrow 10+B=11\\\\\Rightarrow B=11-10\\\\\Rightarrow B=1.

Thus, the required solution is

y(t)=(-2+1\times t)e^{-5t}\\\\\Rightarrow y(t)=(-2+t)e^{-5t}.

6 0
3 years ago
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Order the decimals from least to greatest...
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6 0
3 years ago
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Nickel-63 has half-life of about 96 years. after 336 years,about how many milligrams of an 800 mg sample will remain?
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F=ar^t if the half-life is 96 years...

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5 0
3 years ago
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