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lions [1.4K]
3 years ago
8

Toyota campy sold 20,600 in 2008. percent change is 1.2%. how many sold in 2009?

Mathematics
1 answer:
Artyom0805 [142]3 years ago
6 0
So 20600=100%
in 2009=1.2% change from 100%

not specific whether percent change is up or down so solve fo r1.2%
d
find 1.2% of 100
percent means parts out of 1001.2%=1.2/100=0.12/10=0.012/1=0.012

'of' means multiply
20600 times 0.012=276.2

the change is 276.2

20600-276.2=20352.8
20600+276.2=20847.2
you must round down since you can't sell 0.2 or 0.8 of a car

the sold either 20,352 or 20,847 cars in 2009

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What is this can anyone help me
Solnce55 [7]

g(x) = (x - 3)² - 4

given f(x) is shifted a units horizontally to the right → f(x) → f(x - a)

If f(x) is shifted by a units horizontally left → f(x) → f(x + a)

If f(x) is shifted by b units vertically up → f(x) → f(x) + b

If f(x) is shifted by b units vertically down → f(x) → f(x) - b

The vertex (0, 0) of f(x) has shifted to (3, -4)

that is 3 units horizontally to the right and 4 units vertically down

hence g(x) = (x - 3)² - 4


4 0
4 years ago
Una lancha que viaja a 10 m/s pasa por debajo de un puente 3 segundos después que ha pasado un bote que viaja a 7 m/s, ¿después
ExtremeBDS [4]

Answer:

La lancha y el bote se encontrarán a 70 metros de distancia del puente.

Step-by-step explanation:

Sea el punto debajo del puente el punto de referencia y que ambas lanchas se desplazan a velocidad a continuación, las ecuaciones cinemáticas para cada embarcación son presentadas a continuación:

Bote a 7 metros por segundo

x_{A} = x_{o}+v_{A}\cdot t (Ec. 1)

Lancha a 10 metros por segundo

x_{B} = x_{o}+v_{B}\cdot (t-3\,s) (Ec. 2)

Donde:

x_{o} - Posición debajo del puente, medido en metros.

x_{A}, x_{B} - Posición final de cada embarcación, medido en metros.

v_{A}, v_{B} - Velocidad de cada embarcación, medida en metros por segundo.

t - Tiempo, medido en segundos.

Para determinar la posición en la que ambas embarcaciones se encuentran, se debe determinar el instante en que ocurre a partir de la siguiente condición: x_{A} = x_{B}

Igualando (Ec. 1) y (Ec. 2) se tiene que:

v_{A}\cdot t = v_{B}\cdot (t-3\,s)

Ahora despejamos el tiempo:

3\cdot v_{B} = (v_{B}-v_{A})\cdot t

t = \frac{3\cdot v_{B}}{v_{B}-v_{A}}

Si sabemos que v_{B} = 10\,\frac{m}{s} y v_{A} = 7\,\frac{m}{s}, entonces:

t = \frac{3\cdot \left(10\,\frac{m}{s} \right)}{10\,\frac{m}{s}-7\,\frac{m}{s}}

t = 10\,s

Ahora, la posición de encuentro es: (x_{o} = 0\,m, v_{A} = 7\,\frac{m}{s} y t = 10\,s)

x_{A} = 0\,m + \left(7\,\frac{m}{s} \right)\cdot (10\,s)

x_{A} = 70\,m

La lancha y el bote se encontrarán a 70 metros de distancia del puente.

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