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7nadin3 [17]
2 years ago
14

What is the area of the figure? A. 52 cm2 B. 52 cm O C. 130 cm O D. 130 cm2

Mathematics
2 answers:
IgorLugansk [536]2 years ago
5 0

Answer:

B is the answer

Art [367]2 years ago
5 0

Answer:

130 cm^2

Step-by-step explanation:

split the figure into two rectangles, one can be 4x5 rectangle and the other can be the 11x10 rectangle. now multiply.

4x5 = 20

11x10=110

20+110 = 130 cm^2

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olga55 [171]
1 yard = 36 inches
2 yards = 72 inches

1 feet = 12 inches
3 feet = 36 inches

72 - 36 = 36 inches

she still needs 36 inches
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2 years ago
I don't understand how to do this problem
Naya [18.7K]
It is 270 i believe hope it helps
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3 years ago
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3 0
3 years ago
Read 2 more answers
A coconut is a fruit that is made up of a hard outer shell that is spherical. Inside, there is a thick layer of coconut meat tha
Dvinal [7]

Answer:

V = 18816.57 in^3

Step-by-step explanation:

So we need to find the volume of the inner sphere. Try to picture the coconut as a circle with another circle inside of it. We know that the bigger circle is 55 inches wide, and the distance between the circumferences of the circles is 11 inches. that means the inner circle is 55 - 2(11) inches wide. This is because we have 11 inches of coconut meat on either side of the inner circle.

55 - 2(11) = 33 inches

The radius is half of the diameter, so r = 16.5 inches.

Now just use the formaula V = \frac{4}{3}\pi  r^{3}

V = (4/3) * 3.14 * (16.5)^3

V = 18816.57 in^3

<em>Also, just a side note.. this is one big coconut..</em>

7 0
3 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.

6 0
3 years ago
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