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ollegr [7]
2 years ago
6

Amir stands on a balcony and throws a ball to his dog, who is at ground level. The ball's height (in meters above the ground), x

xx seconds after Amir threw it, is modeled by: h(x)=-(x-2)^2+16h(x)=−(x−2) 2 +16h, left parenthesis, x, right parenthesis, equals, minus, left parenthesis, x, minus, 2, right parenthesis, start superscript, 2, end superscript, plus, 16 What is the height of the ball at the time it is thrown?
Mathematics
2 answers:
defon2 years ago
5 0

Answer:

At the moment it is thrown, the height of the ball is 12 meters

Step-by-step explanation:

If the function h(x) = - (x-2) ^ 2 + 16 models the height of the ball x seconds after it is thrown, then by this same function we can know the initial height of the ball at the initial moment.

If we represent the initial time instant as x = 0, then by doing h (0) we will get the initial height, just when the ball is thrown.

h(0) = - (0-2) ^ 2 +16\\ h(0) = - (4) +16\\ h(0) = -4 +16

h (0) = 12 meters.

At the moment it is thrown, the height of the ball is 12 meters

elena55 [62]2 years ago
3 0

Answer:16 meters

Step-by-step explanation:

The height of the ball at the time it is thrown is given by

h

(

0

)

h(0)h, left parenthesis, 0, right parenthesis.

h

(

0

)

=

−

2

(

0

)

+

4

(

0

)

+

1

6

=

0

+

0

+

1

6

=

1

6

h(0)

​

=−2(0)

2

+4(0)+16

=0+0+16

=16

​

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Answer:

\$6,082.70  

Step-by-step explanation:

we know that

The  formula to calculate the depreciated value  is equal to  

V=P(1-r)^{x}  

where  

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P is the original value  

r is the rate of depreciation  in decimal  

x  is Number of Time Periods  

in this problem we have  

P=\$15,300\\r=14.25\%=0.1425\\x=6\ years

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V=15,300(1-0.1425)^{6}  

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2 years ago
La fuerza necesaria para evitar que un auto derrape en una curva varía inversamente al radio de la curva y conjuntamente con el
Vika [28.1K]

Answer:

768 libras de fuerza

Step-by-step explanation:

Tenemos que encontrar la ecuación que los relacione.

F = Fuerza necesaria para evitar que el automóvil patine

r = radio de la curva

w = peso del coche

s = velocidad de los coches

En la pregunta se nos dice:

La fuerza requerida para evitar que un automóvil patine alrededor de una curva varía inversamente con el radio de la curva.

F ∝ 1 / r

Y luego con el peso del auto

F ∝ w

Y el cuadrado de la velocidad del coche

F ∝ s²

Combinando las tres variaciones juntas,

F ∝ 1 / r ∝ w ∝ s²

k = constante de proporcionalidad, por tanto:

F = k × w × s² / r

F = kws² / r

Paso 1

Encuentra k

En la pregunta, se nos dice:

Suponga que 400 libras de fuerza evitan que un automóvil de 1600 libras patine alrededor de una curva con un radio de 800 si viaja a 50 mph.

F = 400 libras

w = 1600 libras

r = 800

s = 50 mph

Tenga en cuenta que desde el

F = kws² / r

400 = k × 1600 × 50² / 800

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k = 400/5000

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Paso 2

¿Cuánta fuerza evitaría que el mismo automóvil patinara en una curva con un radio de 600 si viaja a 60 mph?

F = ?? libras

w = ya que es el mismo carro = 1600 libras

r = 600

s = 60 mph

F = kws² / r

k = 2/25

F = 2/25 × 1600 × 60² / 600

F = 768 libras

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