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Lelechka [254]
3 years ago
13

A big cruise ship dropped anchor off the caribbean island of Antigua. The heavy anchor dropped into the water at a rate of 2.5 m

eters per second. After 45 seconds, the anchor was 40 meters below the waters surface
Mathematics
2 answers:
elena-s [515]3 years ago
8 0

Answer:

72.5 meters

29 seconds

Step-by-step explanation:

hjlf3 years ago
7 0
Ok what’s next? Did you give us the answer?
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Sanya has a piece of land which is in the shape of a rhombus. She wants her one daughter and one son to work on the land and pro
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Answer:

9600 m2

Step-by-step explanation:

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timofeeve [1]

Answer:

Z

ㄲ

Respuesta: se necesitan 15 casos para las 4 lámincis.

El primer 3 indica los 3 clavos para el lado izquierdo de la primera lámina. los números del L...4

indican el número de laminas, y el segundo 3 indica el número de claves para el lado derecho de

cada lamina.

lámina y se calcula

A mumero de casos se obtiene relacionando el número de liminas con la cantidad de clavos per

3 + nuncro deliminas) X 3

Cuilice si patria numérico anterior para encontrar la cantidad de clavos necesarios según el número

b. 6 laminas

10 Sáminas

0.

-se de forma física bien identificados.

5: Leer el contenido de la presente guía, realizar las

FECHA: 06 al 17 septiembre 2021

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E de liminas

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8 0
3 years ago
Uses both multiplication and addition to find the result
krok68 [10]

Answer:

Where's the question or picture?


Step-by-step explanation:


8 0
3 years ago
–5(3 – q) + 4 = 5q – 11
mrs_skeptik [129]

Answer:

0

Step-by-step explanation:

–5(3 – q) + 4 = 5q – 11

-15 +5q + 4 = 5q -11

-11 + 5q = 5q - 11

+5q - 5q = -11 + 11

0=0

6 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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